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IN  MEMORIAM 
FLORIAN  CAJORl 


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Elements  of  Algebra 


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BY 

G.  A.  WENTWORTH,  A.M. 

AUTHOR  OF  A  SERIES  OF  TEXT=-BOOKS  IN  MATHEMATICS 


COMPLETE  EDITION 


BOSTON,  U.S.A. 

GINN   &   COMPANY,  PUBLISHERS 

1896 


41 


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Entered,  according  to  Act  of  Congress,  in  tlie  year  1881,  by 

G.  A.  WENTWORTH 
in  the  Office  of  the  Librarian  of  Congress,  at  Washington. 


PREFACE. 


THE  single  aim  in  writing  this  volume  has  been  to  make  an 
Algebra  which  the  beginner  would  read  with  increasing  in- 
terest, intelligence,  and  power.  The  fact  has  been  kept  constantly 
in  mind  that,  to  accomplish  this  object,  the  several  parts  must  be 
presented  so  distinctly  that  the  pupil  will  be  led  to  feel  that  he 
is  mastering  the  subject.  Originality  in  a  text-book  of  this  kind 
is  not  to  be  expected  or  desired,  and  any  claim  to  usefulness  must 
be  based  upon  the  method  of  treatment  and  upon  the  number 
and  character  of  the  examples.  About  four  thousand  examples 
have  been  selected,  arranged,  and  tested  in  the  recitation-room, 
and  any  found  too  difficult  have  been  excluded  from  the  book. 
The  idea  has  been  to  furnish  a  great  number  of  examples  for 
practice,  but  to  exclude  complicated  problems  that  consume  time 
and  energy  to  little  or  no  purpose. 

In  expressing  the  definitions,  particular  regard  has  been  paid  to 
brevity  and  perspicuity.  The  rules  have  been  deduced  from  pro- 
cesses immediately  preceding,  and  have  been  written,  not  to  be 
committed  to  memory,  but  to  furnish  aids  to  the  student  in  fram- 
ing for  himself  intelligent  statements  of  his  methods.  Each  prin- 
ciple has  been  fully  illustrated,  and  a  sufficient  number  of  problems 
has  been  given  to  fix  it  firmly  in  the  pupil's  mind  before  he  pro- 
ceeds to  another.  Many  examples  have  been  worked  out,  in  order 
to  exhibit  the  best  methods  of  dealing  with  different  classes  of 
problems  and  the  best  arrangement  of  the  work ;  and  such  aid  has 
been  given  in  the  statement  of  problems  as  experience  has  shown 
to  be  necessary  for  the  attainment  of  the  best  results.  General 
demonstrations  have  been  avoided  whenever  a  particular  illustra- 
tion would  serve  the  purpose,  and  the  application  of  the  principle 
to  similar  cases  was  obvious.     The  reason  for  this  course  is,  that 


M.'^ORnR.I 


IV  PREFACE. 

the  pupil  must  become  familiar  with  the  separate  steps  from  par- 
ticular examples,  before  he  is  able  to  follow  them  in  a  general 
demonstration,  and  to  understand  their  logical  connection. 

It  is  presumed  that  pupils  will  have  a  fair  acquaintance  with 
Arithmetic  before  beginning  the  study  of  Algebra ;  and  that  suffi- 
cient time  will  be  afforded  to  learn  the  language  of  Algebra,  and 
to  settle  the  principles  on  which  the  ordinary  processes  of  Algebra 
are  conducted,  before  attacking  the  harder  parts  of  the  book. 
"Make  haste  slowly"  should  be  the  watchword  for  the  early 
chapters. 

It  has  been  found  by  actual  trial  that  a  class  can  accomplish 
the  whole  work  of  this  Algebra  in  a  school  year,  with  one  recita- 
tion a  day  ;  and  that  the  student  will  not  find  it  so  difficult  as  to 
discourage  him,  nor  yet  so  easy  as  to  deprive  him  of  the  rewards 
of  patient  and  successful  labor.  At  least  one-fourth  of  the  year 
is  required  to  reach  the  chapter  on  Fractions;  but,  if  the  first 
hundred  pages  are  thoroughly  mastered,  rapid  and  satisfactory 
progress  will  be  made  in  the  rest  of  the  book. 

Particular  attention  should  be  paid  to  the  chapter  on  Factoring ; 
for  a  thorough  knowledge  of  this  subject  is  requisite  to  success  in 
common  algebraic  work. 

Attention  is  called  to  the  method  of  presenting  Choice  and 
Chance.  The  accomplished  mathematician  may  miss  the  elegance 
of  the  general  method  usually  adopted  in  Algebras ;  but  it  is  be- 
lieved that  this  mode  of  treatment  will  furnish  to  the  average 
student  the  only  way  by  which  he  can  arrive  at  an  understanding 
of  the  principles  underlying  these  difficult  subjects.  In  the 
preparation  of  these  chapters  the  author  has  had  the  assistance 
and  cooperation  of  G.  A.  Hill,  A.M.,  of  Cambridge,  Mass.,  to 
whom  he  gratefully  acknowledges  his  obligation. 

The  chapter  on  the  General  Theory  of  Equations  has  been  con- 
tributed by  Professor  H.  A.  Howe  of  Denver  University,  Colorado. 

The  materials  for  this  Algebra  have  been  obtained  from  Eng- 
lish, German,  and  French  sources.  To  avoid  trespassing  upon 
the  works  of  recent  American  authors,  no  American  text-book 
has  been  consulted. 


PREFACE.  V 

The  author  returns  his  sincere  thanks  for  assistance  to  Rev.  Dr. 
Thomas  Hill ;  to  Professors  Samuel  Hart  of  Hartford,  Ct. ;  C.  H. 
Judson  of  Greenville,  S.C. ;  O.  S.  Westcott  of  Racine,  Wis. ;  G.  B. 
Halsted  of  Princeton,  N.  J. ;  M.  W.  Humphreys  of  Nashville, 
Tenn. ;  W.  LeConte  Stevens  of  New  York,  N.Y. ;  G.  W.  Bailey 
of  New  York,  N.Y. ;  Robert  A.  Benton  of  Concord,  N.H. ;  and  to 
Dr.  D.  F.  Wells  of  Exeter. 

There  will  be  two  editions  of  the  Algebra :  one  of  350  pages, 
designed  for  high  schools  and  academies,  will  contain  an  ample 
amount  to  meet  the  requirements  for  admission  to  any  coUege ; 
the  other  will  consist  of  the  Elementary  part  and  about  150  pages 
more,  and  will  include  the  subjects  usually  taught  in  colleges. 

Answers  to  the  problems  are  bound  separately  in  paper  covers, 
and  will  be  furnished  free  to  pupils  when  teachers  apply  to  the 
publishers  for  them. 

Any  corrections  or  suggestions  relating  to  the  work  will  be 
thankfully  received. 

G.  A.  WENTWORTH. 

Exeter,  N.H., 
May,  1881. 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from. 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementswentOOwentrich 


CONTENTS. 


CHAPTER  I.     Definitions: 

Quantity  and  number,  1 ;  numbers,  2 ;  algebraic  numbers,  4 ; 
factors  and  powers,  7  ;  algebraic  symbols,  9 ;  algebraic  expres- 
sions, 10  ;  axioms,  11 ;  exercise  in  algebraic  notation,  13 ;  simple 
problems,  14. 

CHAPTER  11.     Addition  and  Subtraction: 

Addition  of  algebraic  numbers,  16  ;  addition  of  monomials,  17  ; 
addition  of  polynomials,  19;  subtraction  of  algebraic  numbers, 
20  ;  subtraction  of  monomials,  21 ;  subtraction  of  polynomials,  22  ; 
parentheses,  25 ;  simplifying  algebraic  expressions  by  removing 
parentheses,  26  ;  introduction  of  parentheses,  27. 

CHAPTER  III.     Multiplication: 

Multiplication  of  algebraic  numbers,  28 ;  law  of  signs  in  mul- 
tiplication, 29;  multiplication  of  monomials  by  monomials,  30; 
the  product  of  two  or  more  powers  of  a  number,  30 ;  multiplica- 
tion of  polynomials  by  monomials,  31 ;  multiplication  of  polyno- 
mials by  polynomials,  32  ;  special  cases  of  multiplication ;  square 
of  the  sum  of  two  numbers  ;  the  square  of  the  difference  of  two 
numbei-s ;  the  product  of  the  sum  and  difference  of  two  numbers, 
37  ;  square  of  a  trinomial,  39  ;  the  product  of  two  binomials  of  the 
form  x-{-  a  and  a:  +  6,  40  ;  the  powers  of  binomials  of  the  form 
a  ±  6,  42. 

CHAPTER  IV.     Division: 

Division  of  algebraic  numbers,  44 ;  division  of  monomials  by 
monomials,  46 ;  division  of  powers  of  a  number,  46 ;  division  of 
polynomials  by  monomials,  48 ;  division  of  polynomials  by  poly- 
nomials, 49  ;  use  of  parentheses  in  division,  53 ;  special  cases  of 
division  ;  the  difference  of  two  equal  odd  powers  of  two  numbers 
divisible  by  the  difference  of  the  numbers,  54 ;  the  sum  of  two 


VIU  ALGEBRA. 

equal  odd  powers  of  two  numbers  divisible  by  the  sum  of  the  num- 
bers, 55  ;  the  difference  of  two  equal  even  powers  of  two  numbers 
divisible  by  the  difference  and  by  the  sum  of  the  numbers,  56 ; 
the  sum  of  two  equal  even  powers  of  two  numbers  when  each 
exponent  is  the  product  of  an  odd  and  an  even  factor  divisible  by 
the  sum  of  the  powers  expressed  by  the  even  factor,  57  ;  general 
definitions  of  addition,  subtraction,  multiplication,  and  division,  58. 

CHAPTER  V.     Simple  Equations  : 

Definitions,  59  ;  transposition  of  terms,  60  ;  solution  of  simple 
equations,  60  ;  problems  in  simple  equations,  62. 

CHAPTER  VI.     Factors: 

Case  in  which  all  the  terms  of  an  expression  have  a  common 
simple  factor,  68  ;  case  in  which  the  terms  of  an  expression  can  be 
so  arranged  as  to  show  a  common  compound  factor,  69  ;  resolution 
into  binomial  factors  of  trinomials  of  the  form  of  x^  +  (a+6)  x  +  a6, 
70  ;  of  the  form  of  x2— {a+6)x+a6,  71 ;  of  the  form  of  x2+(a— 6)x 
—  a6,  72  ;  of  the  form  of  x"^  —  (a  —  6)  x  —  a6,  73  ;  of  the  form  of 
x2  +  2ax  +  a?-^  74  ;  of  the  form  of  x^  —  2ax  +  a^,  75  ;  resolution 
of  expressions  of  the  form  of  two  squares  with  the  negative  sign 
between  them,  76 ;  resolution  of  the  difference  of  two  equal  odd 
powers,  78 ;  resolution  of  the  sum  of  two  equal  odd  powers,  78 ; 
resolution  of  the  sum  of  two  equal  even  powers,  when  possible, 
79  ;  resolution  of  trinomials  of  the  form  of  x*  +  x^  ]p-  +  y^^  80  ;  of 
the  form  of  2x2  +  5ax  +  2a2,  81 ;  resolution  of  polynomials  which 
are  perfect  powers,  82  ;  resolution  of  polynomials  composed  of  two 
trinomial  factors,  83  ;  resolution  of  polynomials  when  a  compound 
factor  of  the  first  three  terms  is  also  a  factor  of  the  remaining 
terms,  84. 

CHAPTER  VII.     Common  Factors  and  Multiples  : 

Highest  common  factor,  88 ;  method  of  finding  the  highest 
common  factor  by  inspection,  88 ;  method  of  finding  the  highest 
common  factor  by  division,  90  ;  principles  upon  which  this  method 
depends,  90  ;  this  method  of  use  only  to  determine  the  compound 
factor  of  the  highest  common  factor,  92  ;  modifications  of  this 
piethod  required,  93  ;  lowest  common  multiple,  98 ;  method  of 
finding  the  lowest  common  multiple  by  inspection,  98  ;  method 
of  finding  the  lowest  common  multiple  by  division,  100. 


CONTENTS.  IX 

CHAPTER  VIII.     Fractions: 

Reduction  of  fractions  to  their  lowest  terms,  103 ;  reduction 
of  fractions  to  integral  or  mixed  expressions,  106 ;  reduction  of 
mixed  expressions  to  the  form  of  fractions,  107  ;  reduction  of  frac- 
tions to  the  lowest  common  denominator,  110 ;  addition  and 
subtraction  of  fractions,  112  ;  multiplication  of  fractions,  120 ; 
division  of  fractions,  122  ;  complex  fractions,  124. 

CHAPTER  IX.      Fractional  Equations: 

Reduction  of  fractional  equations,  130  ;  reduction  of  literal 
equations,  134. 

CHAPTER  X.     Problems  Producing  Fractional  Equations,  137. 

CHAPTER  XI.  Simultaneous  Equations  of  the  First  Degree, 
151 ;  elimination  by  addition  or  subtraction,  152  ;  elimination  by 
substitution,  154 ;  elimination  by  comparison,  156 ;  literal  simul- 
taneous equations,  159. 

CHAPTER  XII.  Problems  Producing  Simultaneous  Equations, 
106. 

CHAPTER  XIII.      Involution  and  Evolution: 

Powers  of  simple  expressions,  181  ;  law  of  exponents,  181  ; 
powers  of  a  binomial  when  the  terms  of  the  binomial  have  coeffi- 
cients or  exponents,  182  ;  powers  of  polynomials  by  the  binomial 
method,  182  ;  roots  of  simple  expressions,  184 ;  imaginary  roots, 
184  ;  square  roots  of  compound  expressions,  186  ;  square  roots  of 
arithmetical  numbers,  188 ;  cube  roots  of  compound  expressions, 
190  ;  cube  roots  of  arithmetical  numbers,  193. 

CHAPTER  XIV.      Quadratic  Equations: 

Pure  quadratic  equations,  196 ;  affected  quadratic  equations, 
198 ;  literal  quadratic  equations,  203 ;  resolution  of  quadratic 
equations  by  inspection,  200  ;  number  of  roots  of  an  equation, 
208  ;  formation  of  equations  when  the  roots  are  known,  209  ; 
determination  of  the  character  of  the  roots  of  an  equation  by  in- 
spection, 209  ;  determination  of  the  maximum  or  minimum  value 


X  ALGEBRA. 

of  a  quadratic  expression,  211 ;  higher  equations  which  can  be 
solved  by  completing  the  square,  212  ;  problems  involving  quad- 
ratics, 214. 

CHAPTER  XV.      Simultaneous  Quadratic  Equations: 

Solution  when  the  value  of  one  of  the  unknown  numbers  can 
be  found  in  terms  of  the  other,  219  ;  when  each  of  the  equations  is 
homogeneous  and  of  the  second  degree,  222  ;  when  the  equations 
are  symmetrical  with  respect  to  the  unknown  numbers,  223 ; 
problems  producing  simultaneous  quadratics,  226. 

CHAPTER  XVI.      Simple  Indeterminate  Equations: 

The  values  of  the  unknown  numbers  dependent  upon  each 
other,  228 ;  method  of  solving  an  indeterminate  equation  in  posi- 
tive integers,  228. 

CHAPTER  XVII.      Inequalities: 

Fundamental  proposition,  234 ;  an  inequality  reversed  by 
changing  the  signs,  234. 

CHAPTER  XVIII.     Theory  op  Exponents  : 

Fractional  and  negative  exponents,  236 ;  the  meaning  of  a 
fractional  exponent,  237  ;  the  meaning  of  a  negative  exponent, 
237  ;  laws  which  apply  to  positive  integral  exponents  apply  also 
to  fractional  and  negative  exponents,  238  j  exercise  with  mono- 
mials having  fractional  and  negative  exponents,  239 ;  exercises 
with  polynomials  having  fractional  and  negative  exponents,  240  ; 
radical  expressions,  242 ;  reduction  of  surds  to  their  simplest 
forms,  243  ;  comparing  surds  of  the  same  order,  244 ;  comparing 
surds  of  different  orders,  245 ;  addition  and  subtraction  of  surds, 
247  ;  expansion  of  binomials  when  the  terms  are  radical  expres- 
sions, 249  ;  rationalization  of  the  denominator  of  a  radical  expres- 
sion, 249 ;  imaginary  expressions,  251 ;  square  root  of  a  binomial 
surd,  252 ;  equations  containing  radicals,  255 ;  solution  of  an 
equation  with  respect  to  an  expression,  257  ;  reciprocal  equations, 
258. 

CHAPTER  XIX.     Logarithms: 

Common  system  of  logarithms,  261  ;  the  characteristic  of  a 
logarithm,  263  ;  the  mantissa  of  a  logarithm,  264  ;  logarithm  of 


CONTENTS.  XI 

a  product,  264  ;  logarithm  of  a  power  and  of  a  root,  265  ;  logarithm 
of  a  quotient,  206  ;  a  table  of  four-place  logarithms,  270  ;  general 
proofs  of  the  laws  of  logarithms,  276 ;  solution  of  an  exponential 
equation  by  logarithms,  277. 

CHAPTER  XX.     Ratio,  Proportion,  and  Variation: 

Ratio,  278 ;  commensurable  and  incommensurable  ratios,  279 ; 
theorems  of  ratio,  281 ;  proportion,  284 ;  theorems  of  proportion, 
284  ;  variation,  292  ;  direct  variation,  293 ;  inverse  variation,  293. 

CHAPTER  XXI.     Series: 

Infinite  series,  299  ;  finite  series,  299 ;  converging  series,  300 ; 
arithmetical  series,  301 ;  geometrical  series,  308 ;  limit  of  the  sum 
of  an  infinite  geometrical  series,  313  ;  harmomcal  series,  314. 

CHAPTER  XXII.      Choice.     Binomial  Theory: 

Fundamental  principle,  317;  arrangements  or  permutations, 
320 ;  number  of  arrangements  of  n  different  elements  taken  all  at 
a  time,  320  ;  number  of  arrangements  of  n  different  elements  taken 
r  at  a  time,  321 ;  number  of  arrangements  of  n  elements  of  which 
p  are  alike,  q  are  alike,  etc.,  324;  number  of  arrangements  of  n 
different  elements  when  repetitions  are  allowed,  325 ;  selections  or 
combinations,  326  ;  number  of  selections  of  r  elements  from  n  dif- 
ferent elements,  326  ;  the  number  of  ways  in  which  p  elements 
can  be  selected  from  p  +  r  different  elements  the  same  as  the  num- 
ber of  ways  in  which  r  elements  can  be  selected,  327 ;  value  of  r 
for  which  the  number  of  selections  of  n  different  elements  taken 
r  at  a  time  is  the  greatest,  330 ;  the  number  of  ways  of  selecting 
r  elements  from  n  different  elements  when  repetitions  are  allowed, 
336 ;  the  number  of  ways  in  which  a  selection  can  be  made  from 
n  different  elements,  337  ;  the  number  of  ways  in  which  a  selec- 
tion can  be  made  from  p  -\-  q  +  r  elements  of  which  p  are  alike, 
q  are  alike,  r  are  alike,  338  ;  proof  of  the  binomial  theorem  when 
the  exponent  is  positive  and  integral,  342  ;  general  formula  for  the 
expansion  of  (a  +  x)»,  343 ;  general  term  of  the  expansion  of  a 
binomial,  344  ;  greatest  term  of  the  expansion  of  a  binomial,  345 ; 
proof  of  the  binomial  theorem  when  the  exponent  is  fractional, 
340  ;  when  negative,  348  ;  applications,  349. 


Xll  ALGEBRA. 

CHAPTER  XXIII.      Chance: 

Theory  of  probabilities,  351 ;  from  several  events  of  which  only 
one  can  happen,  the  chance  that  some  one  of  them  will  happen, 
352  ;  expectation  from  an  uncertain  event,  357  ;  compound  events, 
360;  the  probability  that  two  independent  events  both  happen, 
361 ;  the  probability  of  a  compound  event,  361 ;  the  relative  proba- 
bility of  doubtful  events,  369  ;  the  actual  happening  of  a  doubtful 
event  changes  the  separate  probabilities  of  the  several  ways  of 
happening  in  the  same  ratio,  370. 

CHAPTER  XXIV.      Interest  Formulas: 

Simple  interest,  375 ;  compound  interest,  376 ;  sinking-funds, 
377 ;  annuities,  378 ;  to  find  the  amount  of  an  unpaid  annuity 
when  the  time,  interest,  and  rate  per  cent  are  given,  378 ;  to  find 
the  present  worth  of  an  annuity  when  the  time  it  is  to  continue 
and  the  rate  per  cent  are  given,  378 ;  to  find  the  present  worth 
..  of  an  annuity  that  begins  in  a  given  number  of  years,  when  the 
time  it  is  to  continue  and  the  rate  per  cent  are  given,  379;  to 
find  the  annuity  when  the  present  worth,  the  time,  and  the  rate 
per  cent  are  given,  380 ;  life  insurance,  381 ;  bonds,  381 ;  to  find 
the  rate  of  interest  received  on  an  investment  in  bonds,  when  the 
current  rate  of  interest,  the  market  value  of  the  bonds,  the  rate 
of  interest  they  bear,  and  the  number  of  years  they  run  are  given, 
381  ;  to  find  the  price  that  may  be  paid  for  bonds  bearing  a  given 
rate  of  interest,  and  having  a  given  number  of  years  to  run,  in 
order  that  the  purchaser  may  receive  a  given  rate  of  interest  on 
Ms  investment,  the  current  rate  of  interest  being  known,  382. 


CHAPTER  XXV.      Continued  Fractions: 

The  form  of  a  continued  fraction,  385 ;  any  proper  fraction  in 
its  lowest  terms  may  be  converted  into  a  terminated  continued 
fraction,  385  ;  the  successive  convergents  of  a  continued  fraction 
are  alternately  greater  and  less  than  the  true  value  of  the  given 
fraction,  386  ;  the  law  for  forming  successive  convergents,  387  : 
the  difference  between  two  consecutive  convergents  is  equal  to 
the  reciprocal  of  the  product  of  the  denominators  of  the  two  con- 
vergents, 388  ;   any  convergent  is  in  its  lowest  terms,  388 ;   the 


CONTENTS.  XIU 

successive  convergents  approach  more  and  more  nearly  to  the  true 
value  of  the  fraction,  389 ;  any  convergent  is  nearer  the  true  value 
of  the  continued  fraction  than  any  other  fraction  with  smaller 
denominator,  389  ;  a  quadratic  surd  may  be  expressed  in  the  form 
of  a  non-terminating  continued  fraction,  390 ;  periodic  continued 
fractions,  391  ;  an  exponential  equation  may  be  solved  by  con- 
tinued fractions,  392  ;  problems  in  continued  fractions,  392. 

CHAPTER  XXVI.      Tiieokv  of  Limits: 

Definitions  of  constant,  variable,  and  limit,  394  ;  a  variable 
may  continually  increase  towards  its  limit,  or  continually  decrease 
towards  its  limit,  or  be  sometimes  greater  and  sometimes  less  than 
its  limit,  395  ;  geometrical  representation  of  the  indefinite  approach 
of  a  variable  to  its  limit,  396  ;  if  two  variables  are  equal,  and  are 
so  related  that  a  change  in  one  produces  such  a  change  in  the 
other  that  they  continue  equal,  and  each  approaches  a  limit,  their 
limits  are  equal,  396 ;  if  two  variables  have  a  fixed  ratio,  and  are 
so  related  that  a  change  in  one  produces  such  a  change  in  the 
other  that  they  continue  to  have  this  ratio,  and  each  approaches 
a  limit,  their  limits  are  in  this  same  fixed  ratio,  397  ;  the  limit  of 
the  sum  of  two  or  more  variables  is  the  sum  of  their  respective 
limits,  398  ;  the  limit  of  the  product  of  two  or  more  variables  is 
the  product  of  their  respective  limits,  398  ;  the  limit  of  any  power 
of  a  variable  is  that  power  of  its  limit,  399  ;  the  limit  of  the  quo- 
tient of  two  variables  is  the  quotient  of  their  limits,  399 ;  the 

x"  —  fl" 

limit  of ,  when  x  approaches  a  as  a  limit,  is  na « —  i,  400 ; 

X  —  a 

convergency  of  infinite  series,  400 ;   tests  of  the  convergency  of 

series :  if  the  terms  of  a  series  are  all  positive,  and  the  limit  of  the 

nth  term  is  0,  then  if  the  limit  of  the  ratio  of  the  {n  +  1)  th  term  to 

the  nth  term  is  less  than  1,  the  series  is  convergent,  402 ;  if  the 

terms  of  a  series  are  alternately  positive  and  negative,  and  the 

limit  of  the  nth  term  is  0,  then  when  the  terms  continually  decrease 

the  series  is  convergent,  404. 

CHAPTER  XXVII.     Indeterminate  Coefficients: 

Theorem  of  indeterminate  coefficients,  405  ;  definition  of  an 
identical  equation,  406  ;  method  of  expanding  an  algebraic  fraction 
in  the  form  of  a  series,  40G  ;  method  of  expanding  an  indicated 


XIV  ALGEBKA. 

root  of  an  expression  in  the  form  of  a  series,  407  ;  method  of 
finding  a  fraction  which  when  expanded  will  produce  a  given 
series,  407  ;  recurring  series,  408  ;  reversion  of  series,  408  ;  resolu- 
tion of  a  fraction  into  partial  fractions,  409. 

CHAPTER  XXVIIL      The  Exponential  Theorem: 

To  expand  a^  in  a  series  of  ascending  powers  of  tc,  412  ;  approx- 
imate value  of  the  constant  e,  413  ;  computation  of  logarithms  to 
the  base  e,  414 ;  modulus  of  the  common  system  of  logarithms,  416  ; 
computation  of  common  logarithms,  416 ;  when  the  difference  of 
two  numbers  is  small  in  comparison  with  either  of  them,  the 
difference  of  their  logarithms  is  proportional  to  the  difference  of 
the  numbers,  416. 

CHAPTER  XXIX.      The  Differential  Method: 

To  express  the  (n  +  1)^^  term  of  a  given  series  in  the  first  terms 
of  the  successive  orders  of  differences,  417  ;  to  express  the  sum  of 
n  terms  of  a  given  series  in  the  first  terms  of  the  successive  orders 
of  differences,  418 ;  piles  of  spherical  shot :  in  the  form  of  a  tri- 
angular pyramid,  421 ;  in  the  form  of  a  pyramid  with  square  base, 
421 ;  in  the  form  of  a  pyramid  with  a  base  rectangular  but  not 
square,  422 ;  series  consisting  of  separable  terms,  424  j  interpola- 
tion of  series,  426. 

CHAPTER  XXX.      The  Theory  of  Numbers: 

Systems  of  notation,  428 ;  to  express  any  integral  number  in 
the  scale  of  r,  429  ;  the  common  system  of  notation,  431  ;  if  a 
prime  number  _p  is  a  factor  of  a6,  and  is  not  a  factor  of  a,  it  is  a 
factor  of  &,  432 ;  a  composite  number  can  be  separated  into  only 
one  set  of  prime  factors,  432  ;  a  common  fraction  in  its  lowest 
terms  will  not  produce  a  terminating  decimal  if  its  denominator 
contains  any  prime  factors  except  2  and  5,  433 ;  divisibility  of 
numbers,  434. 

CHAPTER  XXXI.      Imaginary  Numbers: 

Geometrical  representation  of  real,  numbers,  436  ;  geometrical 
representation  of  imaginary  numbers,  437  ;  operations  with  imagi- 
nary numbers,  438 ;  complex  numbers,  440 ;  operations  with 
complex  numbers,  441  ;  imaginary  exponent,  444 ;  formulas 
derived  by  expanding  e«»,  445. 


CONTENTS.  XV 

CHAPTER  XXXII.     Loci  of  Equations: 

Geometrical  representations  of  equations,  447 ;  form  of  the 
equation  whose  locus  is  a  straight  line,  450 ;  an  equation  whose 
locus  is  a  circle,  450  ;  an  equation  whose  locus  is  an  ellipse,  451 ; 
an  equation  whose  locus  is  a  parabola,  451  ;  an  equation  whose 
locus  is  not  a  regular  geometrical  figure,  452  ;  solution  of  two 
simultaneous  equations  by  constructing  their  loci,  452  ;  method  of 
finding  the  roots  of  equations  by  constructing  their  loci,  454. 


CHAPTER  XXXIII.      Equations  in  "General: 

General  equation  of  the  rdh  degree,  457 ;  F{x)  \s  exactly  divi- 
sible by  X  —  a  if  a  is  a  root  of  F  (x)  =  0,  458  ;  if  x  —  a  is  an  exact 
divisor  of  F{x),  a  is  a  root  of  F{x)  =  0,  458;  if  a  is  a  root  of 
F(x)  =  0,  it  is  an  exact  divisor  of  the  absolute  term,  459;  syn- 
thetic division,  460  ;  if  F{x)  =  (x  —  a)  (x  —  6)  (x  —  c),  etc.,  a,  6,  c, 
etc.,  are  the  roots  of  F{x)  =  0,  461  ;  if  ^(x)  =  0  is  of  the  n^A degree, 
it  has  n  roots  and  no  more,  461 ;  the  relations  between  the  coeffi- 
cients and  the  roots  of  an  equation,  463 ;  solutions  by  factoring, 
464 ;  Descartes'  rule  of  signs,  467 ;  when  all  the  roots  of  a  com- 
plete equation  are  real,  the  number  of  positive  roots  equals  the 
number  of  variations  of  sign,  and  the  number  of  negative  roots 
equals  the  number  of  permanences,  468 ;  a  rational  fraction  can- 
not be  a  root  of  /(x)  =  0,  470  ;  imaginary  roots  enter  equations  in 
conjugate  pairs,  471;  if  F{x)  =  0  is  an  odd  degree,  it  has  at  least 
one  real  root,  471 ;  definition  of  a  derivative,  472 ;  to  find  the 
derivative  of  a  simple  expression,  473 ;  to  find  the  derivative  of  a 
compound  expression,  474  ;  the  derivative  of  F{x)  is  the  coefficient 
of  the  first  power  of  h  in  F{x+  h),  475  ;  the  method  of  detecting 
equal  roots  of  F{x)  =  0,  476;  transformation  of  equations,  478; 
the  signs  of  the  roots  of  /(x)  =  0  are  changed  by  changing  the  signs 
of  the  alternate  terms,  beginning  with  the  second,  479 ;  the  roots 
of  /(x)  =  0  are  multiplied  by  m  by  multiplying  the  second  term  by 
m,  the  third  by  m^,  and  so  on,  480 ;  an  equation  is  obtained  with 
roots  the  reciprocals  of  those  of  /  (x)  =  0  by  reversing  the  order 
of  the  coefficients,  480  ;  method  of  transforming  /  (x)  =  0  into  an 
equation  whose  roots  are  greater  or  less  by  A,  481 ;  method  of 
transforming  /(x)  =  0  into  an  equation  whose  second  term  is 
wanting,  482. 


XVI  ALGEBRA. 

CHAPTEE  XXXIV.      Higher  Numerical  Equations: 

The  commensurablQ  roots  of /(x)  =  0,  484  ;  situation  of  incom- 
mensurable roots,  485  ;  if  two  values  of  x  substituted  in  /  (x)  =  0 
give  results  with  unlike  signs,  at  least  one  root  lies  between  the 
values ;  but  if  the  results  have  like  signs,  no  root  or  an  even  num- 
ber of  roots  lies  between  the  values,  485 ;  the  value  of  /(x),  when 
X  =  m,  is  the  remainder  left  from  dividing  f{x)  by  (x  —  m),  486  ; 
method  of  finding  figures  of  the  roots  by  the  principles  of  differ- 
ences^ 489  ;  Sturm's  theorem,  491 ;  Horner's  method  of  finding 
other  figures  of  the  root  when  one  or  more  figures  have  been 
determined,  492  ;  method  of  solving  an  equation  of  which  the 
root  contains  several  figures  in  its  integral  part,  497  ;  extraction 
of  roots  by  Horner's  method,  498 ;  general  directions  for  solving 
higher  numerical  equations,  499  ;  recurring  equations,  501 ;  any 
recurring  equation  is  reducible  to  an  equation  of  half  its  degree, 
504 ;  exponential  equations,  505 ;  Cardan's  method  for  solving 
cubic  equations,  506 ;  trigonometric  solution  of  cubic  equations, 
508. 


ELEMEISTTS    OF  ALGEBRA. 

CHAPTER   I. 
Quantity  and  Number. 

1.  Whatever  may  be  regarded  as  being  made  up  of 
parts  like  the  whole  is  called  a  duantity. 

2.  To  measure  a  quantity  of  any  kind  is  to  find  how 
many  times  it  contains  another  known  q^iantity  of  the 
same  kind. 

3.  A  known  quantity  which  is  adopted  as  a  standard  for 
measuring  quantities  of  the  same  kind  is  called  a  Unit. 
Thus,  the  foot,  the  pound,  the  dollar,  the  day,  are  units 
for  measuring  distance,  weight,  money,  time. 

4.  A  Number  arises  from  the  repetitions  of  the  unit  of 
measure,  and  shows  how  many  times  the  unit  is  contained 
in  the  quantity  measured. 

6.  When  a  quantity  is  measured,  the  result  obtained  is 
expressed  by  prefixing  to  the  name  of  the  unit  the  number 
which  shows  how  many  times  the  unit  is  contained  in  the 
quantity  measured;  and  the  two  combined  denote  a  quan- 
tity expressed  in  units.  Thus,  7  feet,  8  pounds,  9  dollars, 
10  days,  are  quantities  expressed  in  their  respective  units. 


2  ALGEBRA. 

When  a  question  about  a  quantity  includes  the  unit,  the 
answer  is  a  number ;  when  it  does  not  include  the  unit, 
the  answer  is  a  quantity.  Thus,  if  a  man  who  has  fifteen 
bushels  of  wheat  be  asked  how  many  bushels  of  wheat  he 
has,  the  answer  is  the  number,  fifteen ;  if  he  be  asked  hoiv 
much  wheat  he  has,  the  answer  is  the  quantity,  fifteen 
bushels. 

A  number  answers  the  question,  How  many  ?  a  quantity, 
the  question.  How  much  ? 

Numbers. 

6.  The  symbols  which  Arithmetic  employs  to  represent 
numbers  are  the  figures  0,  1,  2,  3,  4,  5,  6,  7,  8,  9.  The 
natural  series  of  numbers  begins  with  0 ;  each  succeeding 
number  is  obtained  by  adding  one  to  the  preceding  number, 
and  the  series  is  infinite. 

7.  Besides  figures,  the  chief  symbols  used  in  Arithmetic 
are  : 

-\-  (read,  plus),  the  sign  of  addition. 

—  (read,  minus),  the  sign  of  subtraction. 

X  (read,  multiplied  by),  the  sign  of  multiplication. 

-f-  (read,  divided  by),  the  sign  of  division. 

=  (read,  is  equal  to),  the  sign  of  equality. 

Exercise.  —  Eead : 

7  +  12  =  19.  8  +  3-5=   20-15  +  1. 

•   '    9—   4=   5.  24  +  6=   10  X    3. 

6X    4  =  24.  14  —  7  +  5=      6X    2. 

48-f-   3  =  16.  9X5  =  180-^   4. 

8.  Any  figure,  or  combination  of  figures,  as  7,  28,  346, 
has  one,  and  only  one,  value.     That  is,  figures  represent 


NUMBERS.  3 

particular  numbers.  But  numbers  possess  many  general 
properties,  which  are  true,  not  only  of  a  particular  number, 
but  of  all  numbers. 

Thus,  the  sum  of  12  and  8  is  20,  and  the  difference 
between  12  and  8  is  4.  Their  sum  added  to  their  differ- 
ence is  24,  which  is  twice  the  greater  number.  Their 
difference  taken  from  their  sum  is  16,  which  is  twice  the 
smaller  number. 

9.  As  this  is  true  of  any  two  numbers,  we  have  this  gen- 
eral property :  The  sum  of  two  numbers  added  to  their  differ- 
ence is  twice  the  greater  number  ;  the  difference  of  two  numbers 
taken  from  their  sum  is  twice  the  smaller  number.     Or, 

1.  (greater  number  -f-  smaller  number)  -\-  (greater  number 

—  smaller  number)  =  twice  greater  number. 

2.  (greater  number  +  smaller  number)  —  (greater  number 

—  smaller  number)  =  twice  smaller  number. 

But  these  statements  may  be  very  much  shortened ;  for, 
as  greater  number  and  smaller  number  may  mean  any  two 
numbers,  two  letters,  as  a  and  b,  may  be  used  to  represent 
them;  and  2a  may  represent  twice  the  greater,  and  2b 
twice  the  smaller.     Then  these  statements  become  : 

1.  {a-{-h)-\-{a  —  b)=^2a. 

2.  la-{-h)  —  {a  —  b)=2b. 

In  studying  the  general  properties  of  numbers,  letters 
may  represent  any  numerical  values  consistent  with  the 
conditions  of  the  problem. 

10.  It  is  also  convenient  to  use  letters  to  denote  numbers 
which  are  unknown,  and  which  are  to  be  found  from  certain 
given  relations  to  other  known  numbers. 


*  ALGEBRA. 

Thus,  the  solution  of  the  problem,  "Find  two  numbers 
such  that,  when  the  greater  is  divided  by  the  less,  the 
quotient  is  4,  and  the  remainder  3 ;  and  when  the  sum  of 
the  two  numbers  is  increased  by  38,  and  the  result  divided 
by  the  greater  of  the  two  numbers,  the  quotient  is  2  and 
the  remainder  2,"  is  much  simplified  by  the  use  of  letters 
to  represent  the  unknown  numbers. 

11.  The  science  which  employs  letters  in  reasoning  about 
numbers,  either  to  discover  their  general  properties,  or  to 
find  the  value  of  an  unknown  number  from  its  relations 
to  known  numbers,  is  called  Algebra. 

Algebraic  Numbers. 

12.  There  are  quantities  which  stand  to  each  other  in 
such  opposite  relations  that,  when  we  combine  them,  they 
cancel  each  other  entirely  or  in  part.  Thus,  six  dollars 
gain  and  six  dollars  loss  just  cancel  each  other ;  but  ten 
dollars  gain  and  six  dollars  loss  cancel  each  other  only  in 
part.  For  the  six  dollars  loss  will  cancel  six  dollars  of  the 
gain  and  will  leave  four  dollars. 

An  opposition  of  this  kind  exists  in  assets  and  dehts,  in 
income  and  outlay,  in  motion  forwards  and  backwards,  in 
motion  to  the  right  and  to  the  left,  in  time  before  and  after 
a  fixed  date,  in  the  degrees  above  and  below  zero  on  a 
thermometer. 

From  this  relation  of  quantities  a  question  often  arises 
which  is  not  considered  in  Arithmetic ;  namely,  the  sub- 
tracting of  a  greater  number  from  a  smaller.  This  cannot 
be  done  in  Arithmetic,  for  the  real  nature  of  subtraction 
consists  in  counting  backwards,  along  the  natural  series  of 
numbers.  If  we  wish  to  subtract  four  from  six,  we  start 
at  six  in  the  natural  series,  count  four  units  backwards,  and 


ALGEBRAIC    NUMBERS.  6 

arrive  at  two,  the  difference  sought.  If  we  subtract  six 
from  six,  we  start  at  six  in  the  natural  series,  count  six 
units  backwards,  and  arrive  at  zero.  If  we  try  to  subtract 
nine  from  six,  we  cannot  do  it,  because,  when  we  have 
counted  backwards  as  far  as  zero,  the  natural  series  of 
numbers  comes  to  an  end. 

13.  In  order  to  subtract  a  greater  number  from  a  smaller 
it  is  necessary  to  assume  a  new  series  of  numbers,  beginning 
at  zero  and  extending  to  the  left  of  zero.  The  series  to  the 
left  of  zero  must  ascend  from  zero  by  the  repetition  of  the 
unit,  precisely  like  the  natural  series  to  the  riglit  of  zero ; 
and  the  ojjposition  between  the  right-hand  series  and  the 
left-hand  series  must  be  clearly  marked.  This  opposition 
is  indicated  by  calling  every  number  in  the  right-hand 
series  a  positive  number,  and  prefixing  to  it,  when  written, 
the  sign  -f ;  and  by  calling  every  number  in  the  left-hand 
series  a  negative  number,  and  prefixing  to  it  the  sign  — . 
The  two  series  of  numbers  will  be  written  thus  : 


-4,  -3,  -2,  -1,      0,  +1,  +2,  +3,  -t-4, 

I        I        I        I        I        I        I        I        I 

If,  now,  we  wish  to  subtract  9  from  6,  we  begin  at  6  in 
the  positive  series,  count  nine  units  in  the  negative  direction 
(to  the  left),  and  arrive  at  — 3  in  the  negative  series.  That 
is  6  —  9  =  — 3. 

The  result  obtained  by  subtracting  a  greater  number 
from  a  less,  when  both  are  positive,  is  always  a  negative 
number. 

If  a  and  b  represent  any  two  numbers  of  the  positive 
series,  the  expression  a  —  b  will  denote  a  positive  number 
when  a  is  greater  than  b ;  will  be  equal  to  zero  when  a  is 
equal  to  b ;  will  denote  a  negative  number  when  a  is  less 
than  b. 

If  we  wish  to  add  9  to  — 6,  we  begin  at  — 6,  in  the 


6  ALGEBRA. 

negative  series,  count  nine  units  in  the  positive  direction 
(to  the  right),  and  arrive  at  +  3,  in  the  positive  series. 

We  may  illustrate  the  use  of  positive  and  negative  num- 
bers as  follows  : 

-5         0  8  20 

\ 1 \ 1 

DA  C 

Suppose  a  person  starting  at  A  walks  20  feet  to  the  right 
of  ^,  and  then  returns  12  feet,  where  will  he  be  ?  Answer: 
at  Cj  a  point  8  feet  to  the  right  of  A.  That  is,  20  feet 
—  12  feet  =  8  feet;  or  20  — 12  =  8. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  20  feet,  where  will  he  be  ?  Answer  :  at  A, 
the  point  from  which  he  started.     That  is,  20  —  20  =  0. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  25  feet,  where  will  he  now  be  ?  Answer  :  at 
D,  a  point  5  feet  to  the  left  of  A.  That  is,  if  we  consider 
distance  measured  in  feet  to  the  left  of  A  as  forming  a 
negative  series  of  numbers,  beginning  at  A,  20  —  25  =  —  5. 
Hence,  the  phrase,  5  feet  to  the  left  of  A,  is  now  expressed 
by  the  negative  number  —  5. 

14.  Numbers  provided  with  the  sign  +  or  —  are  called 
algebraic  numbers.  They  are  unknown  in  Arithmetic,  but 
play  a  very  important  part  in  Algebra.  In  contradistinc- 
tion, numbers  not  affected  by  the  signs  -}-  or  —  are  termed 
absolute  numbers. 

16.  Every  algebraic  number,  as  +4  or  — 4,  consists  of 
a  sign  +  or  —  and  the  absolute  value  of  the  number  ;  in 
this  case  4.  The  sign  shows  whether  the  number  belongs 
to  the  positive  or  negative  series  of  numbers;  the  absolute 
value  shows  what  place  the  number  has  in  the  positive  or 
negative  series. 


FACTORS    AND    POWERS.  • 

16.  "When  no  sign  stands  before  a  number,  the  sign  -f-  is 
always  understood ;  thus,  4  means  the  same  as  -|-  4,  a  means 
the  same  as  +  a.     But  the  sign  —  is  never  omitted. 

17.  Two  numbers  which  have  one  the  sign  +  and  the 
other  the  sign  — ,  are  said  to  have  unlike  signs. 

18.  Two  numbers  which  have  the  same  absolute  values, 
but  unlike  signs,  always  cancel  each  other  when  combined; 
thus  4-4  —  4  =  0,  +a  — a  =  0. 

19.  The  use  of  the  signs  +  ^i^fl  — ?  to  indicate  addition 
and  subtraction,  must  be  carefully  distinguished  from  their 
use  to  indicate  in  which  series,  the  positive  or  the  negative, 
a  given  number  belongs.  In  the  first  sense,  they  are  signs 
of  operations,  and  are  common  both  to  Arithmetic  and  Al- 
gebra. In  the  second  sense,  they  are  signs  of  opposition, 
and  are  employed  in  Algebra  alone. 

Factors  and  Powers. 

20.  When  a  number  consists  of  the  product  of  two  or 
more  numbers,  each  of  these  numbers  is  called  a  factor  of 
the  product.  ^ 

When  these  numbers  are  denoted  by  letters,  the  sign  X 
is  omitted ;  thus,  instead  of  aXh,  we  write  ah ;  instead 
of  a  X  ^  X  c,  we  write  abc. 

The  expression  ahc  must  not  be  confounded  with  a-\-h-\-G\ 
the  first  is  a  product,  the  second  is  a  sum.  If  a  =  2,  ^  =  3, 
c  =  4,  then 

o^c==2x3X4  =  24; 
a-f.^.  +  c  =  2  +  3  +  4=   9. 

21.  Factors  expressed  by  letters  are  called  literal  factors ; 
factors  expressed  by  figures  are  called  numerical  factors. 


8  ALGEBRA. 

22.  A  known  factor  of  a  product  which  is  prefixed  to  an- 
other factor,  to  show  how  many  times  that  factor  is  taken, 
is  called  a  coefficient.  Thus,  in  7  c,  7  is  the  coefficient  of  c; 
in  1  ax,  7  is  the  coefficient  of  ax,  or,  if  a  be  known,  7  a  is 
the  coefficient  of  x.  When  no  numerical  coefficient  occurs 
in  a  product,  1  is  always  understood.  Thus,  ax  means  the 
same  as  lax. 

23.  A  product  consisting  of  two  or  more  equal  factors 
is  called  a  power  of  that  factor. 

24.  The  index  or  exponent  of  a  power  is  a  small  figure  or 
letter  placed  at  the  right  of  a  number,  to  show  how  many 
times  the  number  is  taken  as  a  factor.  Thus,  a\  or  simply 
a,  denotes  that  a  is  taken  once  as  a  factor ;  a^  denotes  that 
a  is  taken  twice  as  a  factor ;  a^  denotes  that  a  is  taken  three 
times  as  a  factor;  and  a"  denotes  that  a  is  taken  n  times  as 
a  factor.  These  are  read  :  the  first  power  of  a ;  the  second 
power  of  a ;  the  third  power  of  a ;  the  nth.  power  of  a. 

a^  is  written  instead  of  aaa. 

a"  is  written  instead  of  aaa,  etc.,  repeated  n  times. 
The  meaning  of  coefficient  and  exponent  must  be  care- 
fully distinguished.     Thus, 

,     4^a  =  a  -}-  a  -{-  a  -\-  a ; 
a^  =  aXaXaXa. 
Ua  =  S,  4a  =  3 +  3  + 3  +  3  =  12. 

a4  =  3X3X3X3  =  81. 

26.  The  second  power  of  a  number  is  generally  called 
the  square  of  that  number;  thus,  a^  is  called  the  square 
of  a,  because  if  a  denote  the  number  of  units  of  length  in 
the  side  of  a  square,  a^  denotes  the  number  of  units  of 
surface  in  the  square. 


ALGEBRAIC    SYMBOLS.  9 

The  third  power  of  a  number  is  generally  called  the  cube 
of  that  number;  thus,  a^  is  called  the  cube  of  a,  because 
if  a  denote  the  number  of  units  of  length  in  the  edge  of  a 
cube,  a^  denotes  the  number  of  units  of  volume  in  the  cube. 


Algebraic  Symbols. 

26.  Known  numbers  in  Algebra  are  denoted  by  figures 
and  by  the  first  letters  of  some  alphabet ;  as  a,  i,  c,  etc. ; 
a',  b'f  c\  read  a  prime^  b  j^rime,  c  prime,  etc. ;  ai,  bi,  Cj,  read 
a  one,  b  one,  c  one. 

Unknown  numbers  are  generally  known  by  the  last 
letters  of  some  alphabet ;   as  x,  y,  z,  x',  y\  z',  etc. 

27.  The  symbols  of  operations  are  the  same  in  Algebra 
as  in  Arithmetic.  One  point  of  difference,  however,  must 
be  carefully  observed.  When  a  symbol  of  operation  is  omit- 
ted in  the  notation  of  Arithmetic,  it  is  always  the  symbol 
of  addition :  but  when  a  symbol  of  operation  is  omitted  in 
the  notation  of  Algebra,  it  is  always  the  symbol  of  mul- 
tiplication.    Thus,  456  means  400  4-50  +  6,  but  4  ab  means 

4  X  a  X  ^ ;  4|  means  4  +  ^,  but  4-  means  4  X  -• 

28.  The  symbols  of  relation  are  =,  >,  <,  which  stand 
for  the  words  "is  equal  to,"  "is  greater  than,"  and  "is 
less  than,"  respectively. 

29.  The  symbols  of  aggregation  are  the  bar,  I  ;  the  vin- 
culum,   ;   the  parenthesis,   (  );  the  bracket,  [];   and 

X 

the  brace,  {  }.     Thus,  each  of  the  expressions,    , 


^  +  2/, 


(ic+2/)>  r^+2/]»  {^+y}>  signifies  that  x-]ry  is  to  be  treated 
as  a  single  number. 


10  ALGEBRA. 

30.   The  symbols  of  continuation  are  dots, ,  or  dashes, 

,  and  are  read,  '^  and  so  on." 


31.  The  symbol  of  deduction  is  .♦. ,  and  is  read  "hence," 
or  "therefore." 

Algebraic  Expressions. 

32.  An  algebraic  expression  is  any  number  written  in 
algebraic  symbols.  Thus,  8  c  is  the  algebraic  expression 
for  8  times  the  number  denoted  by  c. 

If  an  algebraic  expression  contains  only  integral  forms, 
that  is,  contains  no  letter  in  the  denominator  of  any  of  its 
terms,  it  is  called  an  integral  expression. 

33.  A  term  is  an  algebraic  expression  the  parts  of  which 
are  not  separated  by  the  sign  of  addition  or  subtraction. 
Thus,  3aZ»,  5xy,  8ab-^5xi/  are  terms. 

34.  A  monomial  or  simple  expression  is  an  expression 

which  contains  only  one  term. 

35.  A  polynomial  or  compound  expression  is  an  expres- 
sion which  contains  two  or  more  terms.  A  binomial  is  a 
polynomial  of  two  terms.  A  trinomial  is  a  polynomial  of 
three  terms. 

36.  Like  terms  are  terms  which  have  the  same  letters, 
and  the  corresponding  letters  affected  by  the  same  expo- 
nents. Thus,  Ta^cx^  and  —5a^Gx^  are  like  terms;  but 
1  ahx^  and  — ac^x^  are  unlike  terms. 

37.  The  dimensions  of  a  term  are  its  literal  factors. 

^  ,38.  The  degree  of  a  term  is  equal  to  the  number  of  its 
'dimensions,  and  is  found  by  taking  the  sum  of  the  expo- 
nents of  its  literal  factors.  Thus,  'dxy  is  of  the  second 
degree,  and  bx^yz^  is  of  the  sixth  degree. 


AXIOMS.  11 

39.  A  polynomial  is  said  to  be  homogeneous  when  all  its 
terms  are  of  the  same  degree.  Thus,  1  x^  —  bx^y-\rxyz\^ 
homogeneous,  for  each  term  is  of  the  third  degree. 

40.  A  polynomial  is  said  to  be  arranged  according  to  the 
powers  of  some  letter  when  the  exponents  of  that  letter 
either  descend  or  ascend  in  order  of  magnitude.  Thus,  dax^ 
—  ^hx^  —  Q>ax-\-^h  is  arranged  according  to  the  descend- 
ing powers  of  x,  and  8^  —  ^ax  —  ^hx^-\-^ax^  is  arranged 
according  to  the  ascending  powers  of  x. 

41.  The  numerical  value  of  an  algebraic  expression  is  the 
number  obtained  by  giving  a  particular  value  to  each  letter, 
and  then  performing  the  operations  indicated. 

42.  Two  numbers  are  reciprocals  of  each  other  when  their 
product  is  equal  to  unity.     Thus,  a  and  -  are  reciprocals. 


Axioms. 

43.   1.    Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other. 

2.  If  equal  numbers  be  added   to  equal   numbers,  the 
sums  will  be  equal. 

3.  If  equal  numbers  be  subtracted  from  equal  numbers, 
the  remainders  will  be  equal. 

4.  If  equal  numbers  be  multiplied  into  equal  numbers, 
the  products  will  be  equal. 

5.  If  equal  numbers  be  divided  by  equal  numbers,  the 
quotients  will  be  equal. 

6.  If  the  same  number  be  both  added  to  and  subtracted 
from  another,  the  value  of  the  latter  will  not  be  altered. 

7.  If  a  number  be  both  multiplied  and  divided  by  an- 
other, the  value  of  the  former  will  not  be  altered. 


12  ALGEBRA. 

Exercise  1. 

If  a  =  l,  b  =  2,  c  =  3,  d  =  4.,  e  =  5,  f==0,  find  the  nu- 
merical values  of  the  following  expressions  : 

n     o    _Lo/iQ        o^  A     4:ac      SZ'c      5cd 

1.  9a-\-2o-i-3c  —  Zj.  4.    — — | 

ode 

2.  4.e-3a-3b  +  5c.  5.    Te  +  bcd-^- 

3.  Sabc  —  bcd-\-9cde  —  def.     6.    abc^-^-bcd^— dea^-{-f. 

7.  e'  +  6e%'-i-b*  —  4re'b  —  Aeb'. 

8.    i:  ' 


9.    — 


10. 


'      c^- 

-b'            e' 

11. 

b'^  +  d' 

b^^d''  —  bd 

12. 

e'=  —  d' 

d'  —  b'^  *~*    e''-\-ed-^d^ 


In  simplifying  compound  expressions,  each  term  must  be 
reduced  to  its  simplest  form  before  the  operations  of  addi- 
tion and  subtraction  are  performed. 

In  simplifying  a  term,  omit  the  sign  X  when  possible, 
and  write  indicated  divisions  in  the  form  of  fractions. 

Simplify  the  following  expressions  : 

13.  100  +  80-V-4.  15.    25  +  5X4-10--5. 

14.  75  —  25X2.  16.    24  —  5X4-^10  +  3. 

17.    (24-5)  X  (4 -V- 10 +  3). 

Find  the  numerical  value  of  the  following  expressions,  in 
which  a  =  2,  5  =  10,  x  =  3,  y  =  5  : 

18.  xy  +  4:aX2.  20.    3x  +  7y-v-7 -{- aX  y. 

19.  xy  —  15b-^5.  21.    6b  —  Sy-^2yXb  —  2b. 


ALGEBRAIC    NOTATION.  13 

22.  {U  —  ^y)~-2yXh-\-2b. 

23.  {U  —  ^y)^{2ijXh)-\-2b. 

24.  6b  —  (S?j-^2y)Xb  —  2b. 

25.  6b-^(b  —  y)  —  3x-Ji-bxy^l0a, 

Algebraic  Notation. 

26.  Express  the  sum  of  a  and  b. 

27.  Express  the  double  of  x. 

28.  By  how  much  is  a  greater  than  5  ? 

29.  If  ic  is  a  whole  number,  what  is  the  next  number  above 

it? 

30.  Write  five  numbers  in  order  of  magnitude,  so  that  x 

shall  be  the  middle  number. 

31.  What  is  the  sum  of  ic  +  ^  +  ^  + written  a  times  ? 

32.  If  the  product  is  xy  and  the  multiplier  x,  what  is  the 

multiplicand  ? 

33.  A  man  who  has  a  dollars  spends  b  dollars  ;  how  many 

dollars  has  he  left  ? 

34.  A  regiment  of  men  can  be  drawn  up  in  a  ranks  of  b  men 

each,  and  there  are  c  men  over ;  of  how  many  men 
does  the  regiment  consist  ? 

35.  Write,  the  sum  of  x  and  y  divided  by  c  is  equal  to  the 

product  of  a,  b,  and  m,  diminished  by  six  times  c, 
and  increased  by  tlie  quotient  of  a  divided  by  the 
sum  of  X  and  y. 

36.  Write,  six  times  the  square  of  7i,  divided  by  m  minus  a, 

increased  by  five  b  into  the  expression  c  plus  d 
minus  a. 

37.  Write,  four  times  the  fourth  power  of  a,  diminished  by 

five  times  the  square  of  a  into  the  square  of  b,  and 
increased  by  three  times  the  fourth  power  of  b. 


14  ALGEBRA. 


Exercise  2. 

That  the  beginner  may  see  how  Algebra  is  employed  in 
the  solution  of  problems,  the  following  simple  exercises 
are  introduced : 

1.  John  and  James  together  have  $6.     James  has  twice 

as  much  as  John.     How  much  has  each  ? 

Let  X  denote  the  number  of  dollars  John  has. 

Then  2x  =  the  number  of  dollars  James  has, 

and  x  +  2x  =  the  number  of  dollars  both  have. 

But  G  =  the  number  of  dollars  both  have  ; 

.'.x  +  2x  =  6, 
or  3x=6, 

and  x=  2. 

Therefore,  John  has  ^2,  and  James  has  $4. 

2.  A  stick  of  timber  40  feet  long  is  sawed  in  two,  so  that 

one  part  is  two-thirds  as  long  as  the  other.    Eequired 
the  length  of  each  part. 

Let  3x  denote  the  number  of  feet  in  the  longer  part. 

Then  2x  =  the  number  of  feet  in  the  shorter  part, 

and         3  X  +  2  X  =  the  number  of  feet  in  both  together. 

But  40  =  the  number  of  feet  in  both  together ; 

.-.  3x+2x  =  40, 
or  5x=  40, 

and  x  =  8. 

Therefore,  the  longer  part,  or  Sx,  is  24  feet  long;  and 
the  shorter,  or  2x,  is  16  feet. 

Note.  The  unit  of  the  quantity  sought  is  always  given,  and  only 
the  number  of  such  units  is  required.  Therefore  x  must  never  be  put 
for  money ^  lengthy  tvne,  weight,  etc.,  but  always  for  the  required 
number  of  specified  units  of  money,  length,  time,  weight,  etc. 

The  beginner  should  give  particular  attention  to  this  caution. 


I>I10BLEMS.  16 

3.  The  greater  of  two  numbers  is  six  times  the  smaller, 

'  and  their  sum  is  35.     Required  the  numbers. 

4.  Thomas  had  75  cents.     After  spending  a  part  of  his 

money,  he  found  he  had  twice  as  much  left  as  he 
had  spent.     How  much  had  he  spent  ? 

5.  A  tree  75  feet  high  was  broken,  so  that  the  part  broken 

off  was  four  times  the  length  of  the  part  left  standing. 
Required  the  length  of  each  part. 

6.  Four  times  the  smaller  of  two  numbers  is  three  times 

the  greater,  and  their  sum  is  63.  Required  the 
numbers. 

7.  A  farmer  sold  a  sheep,  a  cow,  and  a  horse,  for  $216. 

He  sold  the  cow  for  seven  times  as  much  as  the 
sheep,  and  the  horse  for  four  times  as  much  as  the 
cow.     How  much  did  he  get  for  each  ? 

8.  George  bought  some  apples,  pears,  and  oranges,  for  91 

cents.  He  paid  twice  as  much  for  the  pears  as  for 
the  apples,  and  twice  as  much  for  the  oranges  as 
for  the  pears.  How  much  money  did  he  spend  for 
each  ? 

9.  A  man  bought  a  horse,  wagon,  and  harness,  for  $350. 

He  paid  for  the  horse  four  times  as  much  as  for  tlie 
harness,  and  for  the  wagon  one-half  as  much  as  for 
the  horse.     What  did  he  pay  for  each  ? 

10.  Distribute  $S  among  Thomas,  Richard,  and  Henry,  so 

that  Tliomas  and  Richard  shall  each  have  twice  as 
much  as  Henry. 

11.  Three  men,  A,  B,  and  C,  pay  $1000  taxes.     B  pays  4 

times  as  much  as  A,  and  C  an  amount  equal  to  the 
sum  of  what  the  other  two  pay.  How  much  does 
each  pay  ? 


CHAPTER   II. 

Addition  and  Subtraction. 

44.  An  algebraic  number  which  is  to  be  added  or  sub- 
tracted is  often,  inclosed  in  a  parenthesis,  in  order  that  the 
signs  +  and  —  which  are  used  to  distinguish  positive  and 
negative  numbers  may  not  be  confounded  with  the  +  and 
—  signs  that  denote  the  operations  of  addition  and  subtrac- 
tion. Thus,  -f-  4  +  (—  3)  expresses  the  sum,  and  -f  4  —  (—  3) 
expresses  the  difference,  of  the  numbers  +  4  and  —  3. 

45.  In  order  to  add  two  algebraic  numbers,  we  begin  at 
the  place  in  the  series  which  the  first  numher  occupies,  and 
count,  in  the  directioji  indicated  by  the  sign  of  the  second 
numher,  as  many  units  as  are  equal  to  the  absolute  value 
of  the  second  number.  Thus,  the  sum  of  +  4  +  (+  3)  is 
found  \)y  counting  from  -}~4  three  units  in  the  positive 
direction,  and  is,  therefore,  -f  7  ;  the  sum  of  -j-  4  +  ( —  3)  is 
found  by  counting  from  +4  three  units  in  the  negative 
direction,  and  is,  therefore,  + 1- 

In  like  manner,  the  sum  of  —  4  +  (+  3)  is  —  1,  and  the 
sum  of  —  4  +  (—  3)  is  —  7.     That  is, 

(1)  +4  +  (+3)  =  7;  (3)  -4  +  (+3)  =  -l; 

(2)  +4  +  (-3)  =  l;  (4)  -4  +  (-3)  — 7. 

I.  Therefore,  to  add  two  numbers  with  like  signs,  find 
the  sum  of  their  absolute  values,  and  prefix  the  common  sign 
to  the  sum. 

II.  To  add  two  numbers  with  unlike  signs,  find  the 
difference  of  their  absolute  values,  and  prefix  the  sign  of  the 
greater  num.ber  to  the  difference. 


ADDITION.  17 


Exercise  3. 

1.  +16  +  (-ll)=  3.   +68  +  (-79)  = 

2.  -15  +  (-25)=  4.    -7  +  (+4)  = 

5.  +33  +  (+18)  = 

6.  +378 +  (+709) +  (-592)  = 

7.  A  man  has  $5242  and  owes  $2758.     How  much,  is  he 

worth  ? 

8.  The  First  Punic  War  began  b.c.  264,  and  lasted  23 

years.     When  did  it  end  ? 

9.  Augustus  Caesar  was  born  b.c.  63,  and  lived  77  years. 

When  did  he  die  ? 
10.    A  man  goes  65  steps  forwards,  then  37  steps  backwards, 
then  again  48  steps  forwards.     How  many  steps  did 
he  take  in  all  ?     How  many  steps  is  he  from  where 
he  started  ? 

Addition  op  Monomials. 

46.  If  a  and  b  denote  the  absolute  values  of  any  two 
numbers,  1,  2,  3,  4  (§  45)  become : 

(1)  ^a  +  (+b)  =  a-\-b',  (3)  -a  +  (+^)  =  -a  +  J; 

(2)  ■i-a-{-(-b)  =  a  —  b;  (4)  - a-{-(-b)  =  - a-b. 

Therefore,  to  add  two  terms,  wj'ite  them  one  after  the  other 
with  unchanged  signs. 

It  should  be  noticed  that  the  order  of  the  terms  is  im- 
material. Thus,  -\-a  —  b=^  —  b-\-a.  If  a  =  8  and  b  =  12, 
the  result  in  either  case  is  —  4. 

47.  3a  +  5a^2a-i-6a-\-a  =  17a. 
—  2g  —  Sg  —  c  —  4:C  —  Sc  =  —  1Sc. 

Therefore,  to  add  several  like  terms  which  have  the  same 


18  ALGEBRA. 

sign,  add  the  coefficients,  prefix  the  common  sign,  and  annex 
the  common  symbols. 

48.    la  —  Q>a-\-lla-\-a  —  ^a  —  2a  =  l^a  —  l^a  =  Qa. 

—  3a  —  15a  —  7a-\-14:a  —  2a  =  14:a  —  27a  =  —  13a. 

Therefore,  to  add  several  like  terms  which  have  not  all 
the  same  sign,  find  the  difference  between  the  sum  of  the 
positive  coefficients  and  the  sum.  of  the  negative  coefficients, 
prefix  the  sign  of  the  greater  sum,  and  annex  the  common 


49.  5a  —  2b-]-Sa  =  Sa'-2b. 

—  Sax-\-Sy-\-9ax  —  4:C=^6ax-\-Sj/  —  4  c. 

Therefore,  to  add  terms  which  are  not  all  like  terms, 
combine  the  like  terms,  and  write  down  the  other  terms,  each 
preceded  by  its  proper  sign. 

Exercise  4. 

1.  bab-{-{—^ab)=  6.    1  ab-]-{—5ab)  = 

2.  ^mx-^{—2mx)=  1.    12() my -\-{— 9b my)  = 
Z,    — 13  mng-}-(— 7  mng)=     8.    —  33  aP-\-(-i- 11  ab^)  = 

4.  -5x'  +  (-^Sx')=  9.    ~75xy-\-(+20x7/)  = 

5.  25  m?/ -{-(— IS  7n?f)=       10.    -i-15a^x^  + (—a^x^)  = 

11.  —b^m^^(+7b''m^)  = 

12.  5a  +  (-3^»)  +  (+4a)  +  (-7^)  = 

13.  4.a^c  +  (—  lOxijz)  +  (+  6 a^c)  +  (-  9xyz) 

+  (— 11  a^c)  +  (+  20xyz)  = 

14.  3x^y  +  (-4.ab)-{-(-2mn)-\-(+5x^y) 


ADDITION.  19 


Addition  of  Polynomials. 

60.  Two  or  more  polynomials  are  added  by  adding  their 
separate  terms. 

It  is  convenient  to  arrange  the  terms  in  columns,  so  that 
like  terms  shall  stand  in  the  same  column.     Thus, 

(1)  2a^-3aH-}-4:ab^+    h^  {2)—2x'y              -\-Qif  —  l 

a^-\-4:aH  —  7ab^-2b^  -4.a^y-{-2xif            +5 

—  3a3+    aH  —  ZaV  —  W  6xV                         +2 

2a^-\-2aH-\-&ab^  —  3h^  x^y              —    f 

2a?^^a?h  —U^  -2xV -5 


Exercise  6. 
Add: 

1.  6a  +  3^»  +  c,   Sa  +  SZ'  +  Sc,   «  +  3^»  +  5c. 

2.  la  —  ^b-\rc,    6a  +  3^>  — 5c,    — 12a  +  4c. 

3.  a-\-b  —  c,    b-\-c  —  a,    c-\-a  —  5,    a-\-b  —  c. 

4.  a-\-2b-\-ZG^    2a  —  b  —  2c,    b  —  a  —  c,    c  —  a  —  b. 

5.  a  — 2^»  +  3c  — 4^,    3b  —  4:C-\-od  —  2a, 

5c  —  6d  +  3a  —  U,    7d  —  4:a  +  5b  —  4:C. 

6.  x«-4a;2  +  5x-3,   2x^-7x^  —  7x^-Ux-\-5, 

-x^-i-9x^  +  x-\-S. 

7.  x'~2x'-{-Sx',   x'-]-x'-^x,    4.x'-\-5x% 

2x'  +  Sx-4.,   -Sa^-2x-5. 

8.  a^  +  3ab^  —  SaH  —  b',    2a^-{-5a^  —  6ab^  —  7b^, 

a^'-ab^-\-2b^ 

9.  2ab  —  3ax^-\-2a^x,    12 ab  —  6a^xi- 10 ax", 

ax^  —  Sab  —  5a^x. 


JU  ALGEBRA. 

10.  c*  —  Sc^  +  2c^  —  4:C  +  T,   2c'-\-3c^  +  2c^  +  5c  +  6, 

11.  3x^  —  x7/-{-xz  —  3y^-\-4:2/^  —  z^,  — 5x^  —  xy  —  xz-\-5yz 

6x^  —  6y  —  6z,   4tyz  —  5ijz-\-3z% 
—  4.x^-{-2/  +  3yzi-3z\ 

12.  m*  —  3  m^n  —  6  m\^,   +  ^^^^  "l~  ^^^^  —  ^  ^"^^j 

7  m^Ti^  +  4  m^/i^  —  3  mn*,   —  2  m  V  —  3  7nn'*^  +  4  w,*, 
2mn'^-i-2n^-i-3m^,   — 7i^  +  2m^  +  7m%. 

Subtraction. 

51.  In  order  to  find  the  difference  between  two  algebraic 
numbers,  we  begin  at  the  place  in  the  series  which  the  minu- 
end occupies,  and  count  in  the  direction  opposite  to  that  indi- 
cated hy  the  sign  of  the  subtrahend  as  many  units  as  are 
equal  to  the  absolute  value  of  the  subtrahend. 

Thus,  the  difference  between  +  4  and  +  3  is  found  by- 
counting  from  -}-4  three  units  in  the  negative  direction, 
and  is,  therefore,  +1;  the  difference  between  +4  and  — 3 
is  found  by  counting  from  +  ^  three  units  in  the  positive 
direction,  and  is,  therefore,  +  '^• 

In  like  manner,  the  difference  between  —  4  and  +  3  is 
—  7 ;    the  difference  between  —  4  and  —  3  is  —  1. 

Compare  these  results  with  results  obtained  in  addition : 


+  4  +  (-3)  =  l. 
+  4  +  (+3)  =  7. 
-4  +  (-3)  =  -7. 
-4  +  (+3)  =  -l. 


(1)  +4-(+3)  =  l 

(2)  +4-(-3)  =  7 

(3)  -4-(+3)  =  -7 

(4)  -4-(-3)  =  -l 
Or,     (1)  +4-(+3)  =  +  4+(-3). 

(2)  +4-(-3)  =  +  4  +  (+3). 

(3)  -4-(+3)  =  -4  +  (-3). 

(4)  -4-(-3)=-4  +  (+3). 


SUBTRACTION.  21 

62.  From  (1)  and  (3),  it  is  evident  that  subtracting  a 
positive  number  is  equivalent  to  adding  an  equal  negative 
number. 

From  (2)  and  (4),  it  is  evident  that  subtracting  a  nega- 
tive number  is  equivalent  to  adding  an  equal  positive  number. 

To  subtract,  therefore,  one  algebraic  number  from  an- 
other, change  the  sign  of  the  subtrahend,  and  then  add  the 
subtrahend  to  the  minuend. 


Exercise  6. 

1.  +25-(+16)=  3.   -   31-(+58)  = 

2.  -50-(-25)=  4.    -fl07-(-93)  = 

5.  Rome  was  ruled  by  emperors  from  b.c.  30  to  its  fall, 

A.D.  476.     How  long  did  the  empire  last  ? 

6.  The  continent  of  Europe  lies  between  36"  and  71°  north 

latitude,  and  between  12°  west  and  63°  east  longitude 
(from  Paris).  How  many  degrees  does  it  extend  in 
latitude,  and  how  many  in  longitude  ? 


Subtraction  of  Monomials. 

If  a  and  b  denote  the  absolute  values  of  any  two  num- 
bers, 1,  2,  3,  and  4  (§  51)  become  : 

(1)  -\-a  —  {-^'b)  =  a  —  b.  (3)    —a  —  {-\-b)=  —  a  —  b. 

(2)  -\-a—{—b)  =  a-{-b.  (4)    —  a—{—b)=  — a-\-b. 

To  subtract,  therefore,  one  term  from  another,  change 
the  sign  of  the  term  to  be  subtracted,  and  vrrite  the  terms  one 
after  the  other. 


22  ALGEBRA. 

Exercise  7. 

1.  5x—(—4:x)=  6.  17ax'-(-24.ax')  = 

2.  —Sab—(-\-5ab)=  7.  Ba^'x  — (—3a^x)  = 

3.  3a^>2_(_|_i0a^;2^)^  8.  ~Axf/~  (—5xi/)  = 

4.  157?i2x2  — (— 7??i2a^')=:  9.  8«^_(_3^^)  = 

5.  —7aij—(—3a?/)=  10.  2  aO^  i/ ~  (-}- abi/)  = 

11.  9a^2+(5:r2)-(+8a^2^^  = 

12.  5a;2y— (- 18 a;2^)  +  (— 100^2?/)  = 

13.  17aa;^— (— a:^^)  — (+24aa3«)  = 

14.  —3ab-\-(2mx)  —  (—4:7nx)  = 

15.  3a— (+2Z>)  — (— 4c)  = 

SUBTRACTIOX    OF    POLYNOMIALS. 

53.  When  one  polynomial  is  to  be  subtracted  from  an- 
other, place  its  terms  under  the  like  terms  of  the  other, 
if  they  have  like  terms,  change  the  signs  of  the  subtrahend, 
and  add. 

From  4:X^  —  8x^i/—   xi/-\-2^ 

take  2x^—   x^i/-\-5xif  —  3^ 

Change  the  signs  of  the  subtrahend  and  add  : 

4:X^  —  3x^y—    xi^-\-2y^ 

—  2x^-\-    x^ij  —  bxif--3'if 
2x^  —  2xhj-Qxy''^Bf 

From  a^'x^  +  2  a^x^  —  4  ax* 

take  a''  +  4a3a;2  — 3aV  — 4aic* 

—  a^  —  3a^x^-\-5a^x^ 


SUBTRACTION.  23 

In  the  last  example  we  have  conceived  the  signs  to  be 
changed  without  actually  changing  them.  The  beginner 
should  do  the  examples  by  both  methods  until  he  has  ac- 
quired sufficient  practice,  when  he  should  use  the  second 
method  only. 

Exercise  8. 

1.  YTom6a  —  2b  —  ctsike2a  —  2b  —  3c. 

2.  ¥Tom3a  —  2b-}-3Gtake2a  —  7b  —  c  —  b. 

3.  FromTir^  — 8a;  — 1  take  50-2  — 6.^  +  3. 

4.  From  4a;^  —  3x-^  — 2^2— 7^  +  9 

take  x'  —  2x^  —  2x^-\-7x  —  9. 

5.  YTom2x^  —  2ax-\-3aHakex^  —  ax-\-a\ 

6.  From  x^  —  3x1/  —  i/ -\-  f/z  —  2 z^ 

take  x^-{-2xy-\-5xz  —  3i/'  —  2z\ 

7.  From  a^  —  3  aH  +  3  td>^  —  b^ 

tQke—a^  +  3a^  —  3ab^  +  b^ 

8.  ¥Tomx^  —  r)xi/-\-xx  —  if-\-7yz  +  2z- 

take  x^  —  xf/  —  xz-\-2i/z -^3z^. 

9.  Yvom2ax^-\-3abx  —  Ab^x-\-12b^ 

take  ax^  —  4  abx  -{-bx^  —  5b^x  —  x^. 

10.  From  6x^—7x^i/  +  4:Xi/^  —  2i/^-5x^  +  xf/  —  4:?f-i-2 

take  Sx'-7x'i/  +  xif-i/-\-dx'-xi/  +  6y'  —  4.. 

11.  From   a^  — ^^^"take  4:a^  —  6a%^  +  4:ab^,   and  from  the 

result  take  2a'-4.a='b-{-6a'P-{-4.ab'-2b\ 

12.  Ytoiu  x^  ?f  —  3x^ / -\-  ixi/  —  f  take  —x^-\-2x*f/  —  4:xy* 

—  4//*.     Add  the  same  two  expressions,  and  subtract 
the  former  result  from  the  latter. 

13.  From  a^b^-'aHc  —  Sab'c-ah^  +  abc^-GPc^ 

take  2a''bc  —  5ab^c-\-2abc^  —  5b^(^. 


24  ALGEBRA. 

14.  From    12a  +  Sb  —  5c  —  2d   take    10a  — b -{- 4:6  —  8(1, 

and  show  that  the  result  is  numerically  correct 
when  a  =  6,  b  =  4:,  c  =  l,  d  =  5. 

15.  What  number  must  be  added  to  a  to  make  b ;  and  what 

number  must  be  taken  from  2a^  —  6a%-\-6ab^  —  2b'^ 
to  leave  a^  —  7a^  —  3b^? 

16.  Yrom2x'  —  if  —  2x?/-^znsikex^—?f-\-2xi/—z\ 

17.  From  12ac-{-Scd  —  d  take  — 7ac  — 9c(^  +  8. 

18.  From  —6a^  +  2ab  —  3c^  take  ^a^'-^-Qab  —  4c\ 

19.  From  ^xy  —  4x  —  3y-\-l  take  8xy  — 2x  +  3?/  +  6. 

20.  From  —  a^bc  —  ab'^c  +  abc^  —  abc 

take  a^bc  -\-  ab'^e  —  abc^  -\-  abc. 

21.  From7«2  — 2ic  +  4  take  2x2  +  3x  — 1. 

22.  From  2>x^-\-2xy—  if  take  —x'^  —  3xy-{-  3 ?/^  and  from 

the  remainder  take  3x^-{-4zxy  —  5y\ 

23.  From  ax^  —  %^  take  cic^  —  dy^. 

24.  From  ax-\-bx-{-by-\-  cy  take  ax  —  bx  —  by-\-  cy. 

25.  From  Sx^-f  4ic  — 4y  +  3?/Hake  ox'^  —  3x-\-3y^y^. 

26.  From  a^b^  +  12 abc  —  9 ax^  take  4ab'^  —  ^ acx  +  3 a^x. 

27.  From  a^  —  2ab  -{-  c^  —  3b^  t2,\LQ  2a^ -  2ab  ^3h\ 

28.  From  the  sum  of  the  first  four  of  the  following  expres- 

sions, a^^b''-{-c''-\-d\  d'^-\-b'~-^c\  a^-c'-^b''  —  d\ 
a^  —  b^-Jrc^-{-d%  b'-\-c'+d'  —  a^  take  the  sum  of 
the  last  four. 

29.  From  2x''  —  2y'  —  z^  take  Sy^-\-2x^  —z",  and  from  the 

remainder  take  3z^  —  2if  —  x^. 

30.  From  a^  —  2  a"^ c -\- 3  ac^  take  the  sum  of  ah  —  2a^  +  2ac^ 


PARENTHESES.  26 

PARENTHESES.  , 

64.   From  (§  52),  it  appears  that 

(1)  a  +  (+b)  =  a  +  b 

(2)  a-\-(—b)  =  a  —  b. 

(3)  a  —  (+b)  =  a  —  b. 

(4)  a—(—b)  =  a-\-b. 

The  same  laws  respecting  the  removal  of  parentheses 
hold  true  whether  one  or  more  terms  are  inclosed.  Hence, 
when  an  expression  within  a  parenthesis  is  preceded  by  a 
plus  sign,  the  parenthesis  may  be  removed. 

When  an  expression  within  a  parenthesis  is  preceded  by 
a  minus  sign,  the  parenthesis  may  be  removed  if  the  sign 
of  every  term  within  the  parenthesis  he  changed.     Thus  : 

(1)  a-\-(b  —  c)  =  a-\-b  —  c, 

(2)  a  —  (b  —  c)^a  —  b-\-c. 

56.  Expressions  may  occur  with  more  than  one  paren- 
thesis. These  parentheses  may  be  removed  in  succession 
by  removing  Jirstf  the  innermost  j^a^^^^thesls ;  next,  the 
innermost  of  all  that  remain,  and  so  on.     Thus  : 

(1)    a-{b-(c-d)l 
=  a-{b-c-^dl 
^a  —b-\-  r  —  d. 


(2)    a-[b-\c-\-(d-e-f)U 
=  a-lb-\c+(d-e+f)\l 
=  a-[b-{c^d-e+f}l 
=  a  —  [b  —  c  —  d-\-e  — /], 
=  a  —  b-\-c-\-d  —  e  -\-f 


26  ALGEBRA. 

Exercise  9. 

Simplify   the   following   expressions   by   removing   the 
parentheses  and  combining  like  terms  : 

1.  (a-\-b)^(b-Yc)~(^i  +  c). 

2.  (2a  —  b  —  G)  —  {a  —  2h-\-c). 

3.  (^2x-y)-(2y-z)-(2z-x). 

4.  (a-x-y)-{h-x-\-y)^{c^2y). 

5.  (2x-y^^z)-\-{-x-y-4.z)-{^x-2y-z). 

6.  (3a  —  ^»  +  7c)  —  (2a  +  3^)  —  (oZ*  —  4 1")  +  (3 c  —a). 

7.  l-(l-a)  +  (l  — a  +  a=^)-(l-a  +  a2-a3). 

8.  a—\2h  —  {3G^2h)-a\. 

9.  2a  —  \b  —  {a  —  2b)\. 

10.  3a— J^>  +  (2a  — Z»)  — (a  — ^»)}. 

11.  ta  — [3a  — J4a  — (5a  — 2a)5]. 

12.  2x  +  (?/  — 3^)— J(3a;  — 2?/)  +  ^J+5a;  — (4?/  — 3«). 

13.  J(3a-2^')  +  (4c  — a);  — Ja— (25  — 3a)— 6'J 

+  Ja-(/>-5c-a)^. 

14.  a— [2a+(3a— 4a)]  — oa— J6a  — [(7a+8a)— 9a]J. 

15.  2a—{U-{-2c)  —  [pb  —  {Qc  —  U)-\-5G 

-J2a-(.  +  25)n. 

ps^ae.    a— [2^/+J3c  — 3a— (a  +  Z*)J  +  J2a— (Z^  +  c)^]. 


X  17.    16  — ir  — [7a^— JSa;- (9cc  — 3:r  — 6a;)5].- 


^    18.    2a— [3Z'  +  (2Z'  — c)— 4c  +  J2a  — (3^*- c  — 2^*);]. 
^    19.    a  — [2^'+J3c  — 3a  — (a  +  ^')J  +  2a— (/>  +  36')]. 
y  20.    a— [55-i-^a  — (3c  — 3/^)+2c— (a  — 2^»  — c)J]. 
^-^  -x 


( 


PARENTHESES.  27 


56.  The  rules  for  introducing  parentheses  follow  directly 
from  the  rules  for  removing  them  : 

1.  Any  number  of  terms  of  an  expression  may  be  put  C^— . 
within  a  parenthesis,  and  the  sign  plus  placed  before  the  AJj. 
whole.  *_1— 

2.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  minns  placed  before  the 
whole  ;  provided  the  sign  of  every  term  within  the  paren- 
thesis he  changed. 

It  is  usual  to  prefix  to  the  parenthesis  the  sign  of  the 
first  term  that  is  to  be  inclosed  within  it. 

Exercise  10. 
Express  in  binomials,  and  also  in  trinomials  : 

1.  2a  — 3Z»  — 4c  +  ^/  +  3e  — 2/. 

2.  a  — 2a!  +  4y  — 3,t  — 2^  +  c. 

3.  a'^3a^  —  2a^  —  4:a^-\-a  —  l. 

4.  —^a  —  2h-\-2c-bd  —  e  —  2f. 

5.  ax  —  hjj  —  cz  —  hx-\-  cy-\-  az. 

6.  2x'  —  3x*y-\-4:X^y''  —  6x^if-\-xif  —  2y'. 

7.  Express  each  of  the  above  in  trinomials,  having  the 

last  two  terms  inclosed  by  inner  parentheses. 

Collect  in  parentheses  the  coefficients  of  x,  y,  z,  in 

8.  2'Ax  —  ^ay-\-'ihx  —  Ux  —  2cx  —  ^cy.        < 

9.  alc  —  hx-\-2ay-\-^y-\-4.az  —  Zhz  —  2z.       < 

10.  ax  —  2hy-\-^cz  —  4:hx  —  '^cy-\-az  —  2cx  —  ay-\-A:hz.  V 

11.  12ax-\-12ay-{-Uy  —  12hz  —  lDCX  +  Q^cy-\-Zcz.       > 

12.  2ax  —  "6hy  —  lcz  —  2hx-\-2cx-\-Scz  —  2cx  —  cy  —  cz.   K 


CHAPTER   III. 
Multiplication  of  Algebraic  Numbers. 

67.  The  operation  of  finding  the  sum  of  3  numbers,  each 
equal  to  5,  is  symbolized  by  the  expression,  3  X  5  — 15, 
read,  '^  three  times  five  is  equal  to  fifteen "  ;  or,  by  the 
expression  5  X  3  =  15,  read,  "  five  multiplied  by  three  is 
equal  to  fifteen." 

68.  With  reference  to  this  operation,  this  sum  is  called 
the  product ;  one  of  the  equal  numbers  is  called  the  multi- 
plicand ;  and  the  number  which  shows  how  many  times  the 
multiplicand  is  to  be  taken  is  called  the  multiplier. 

69.  The  multiplier  means  so  many  times.  The  multipli- 
cand can  be  a  positive  or  a  7iegative  number ;  but  the  mul- 
tiplier, when  integral,  means  that  the  multiplicand  is  taken 
so  many  times  to  he  added,  or  to  be  subtracted. 

60.  If  we  have  to  multiply  867  by  98,  we  may  put  the 
multiplier  in  the  form  100—2.  The  100  will  mean  that 
the  multiplicand  is  taken  100  times  to  be  added ;  the  — 2 
will  mean  that  the  multiplicand  is  taken  twice  to  be  sub- 
tracted. 

In  general,  an  integral  multiplier  with  -\-  before  it,  ex- 
pressed or  understood,  means  that  the  multiplicand  is  taken 
so  many  times  to  be  added ;  and  a  multiplier  with  —  before 
it  means  that  the  multiplicand  is  taken  so  many  times  to 
he  subtracted.     Thus, 


MULTIPLICATION.  29 

(1)  +3X(+5)  =  (-f5)  +  (+5)-f-(+5),  or(+15). 

(2)  +3X(-5)  =  (-5)  +  (-5)  +  (-5),  or(-lo). 

(3)  -3X(+5)=-(-f5)-(+5)-(+5),or(-lo). 

(4)  -3X(-5)=-(-5)-(-5)-(-5),  or(+15). 

From  these  four  cases  it  follows,  that,  in  finding  the 
product  of  two  numbers, 

61.  Like  signs  produce  plus ;  unlike  signs,  minus. 

Exercise  11. 

1.  —17X8=  4.   — 18X— 5  = 

2.  -12.8X25=  5.    43  X -6  = 

3.  3.29X5.49=  6.    457X100  = 

7.  (—358  — 417)  X— 79  = 

8.  (7.512 -J- 2.8940  X  (-6.037  + J  13.963;)  = 

62.  The  product  of  more  than  two  factors,  each  preceded 
by  — ,  will  be  positive  or  negative,  according  as  the  number 
of  such  factors  is  even  or  odd.     Thus, 

—  2X-3X— 4=    +6X-4=   —24. 
-2X-3X-4X-5  =  -24X-5  =  +  120. 

9.  13  X  8  X  —  7  = 

10.  — 38X9X— 6  = 

11.  —  20.9X  — i  1  X8  = 

12.  —  78.3  X— 0.57  X  + 1.38  X  — 27.9  = 

13.  —  2.906  X— 2.076  X- 1.49X0.89  = 


ALGEBRA. 


Multiplication  of  Monomials. 

63.  The  product  of  numerical  factors  is  a  new  number 
in  which  no  trace  of  the  original  factors  is  found.  Thus, 
4  X  9  =  36.  But  the  product  of  literal  factors  can  only  be 
expressed  by  writing  them  one  after  the  other.  Thus,  the 
product  of  a  and  h  is  expressed  by  ah ;  the  product  of  ah 
and  cd  is  expressed  by  ahcd. 

64.  If  we  have  to  multiply  5  a  by  —  4  &,  the  factors  will 
give  the  same  result  in  whatever  order  they  are  taken. 
Thus,  5aX— 4^>  =  5X— 4XaX^'  =  — 20  Xa^-^  — 20a&. 

66.  Hence,  to  find  the  product  of  monomials,  annex  the 
literal  factors  to  the  product  of  the  numerical  factors. 

66.  a^  X  a^  =  aaX  aaa  =  aaaaa  =  a^. 

a^  X  a^  X  a^=^aaX  aaaX  aaaa  =  aaaaaaaaa  =  a^. 

It  is  evident  that  the  exponent  of  the  product  is  equal  to 
the  sum  of  the  exponents  of  the  factors.     Hence, 

67.  The  product  of  two  or  more  powers  of  any  numher  is 
that  nurnher  with  an  exponent  equal  to  the  sum  of  the  expo- 
nents of  the  factors. 

Exercise  12. 

1.  -\-aX-\-h=^-\-ah.  6.  —^p  X^m  =  ~2^pm. 

2.  -\-aX—h  =  —  ah,  7.  ^a^X  —  a^  =  —  ^a\ 

3.  —aX-\-h=^  —  ah.  8.  —3aX2a^  =  —  Qa\ 

4.  —aX—h  =  -{-ah.  9.  6<xX— 2a  = 

5.  7aX5h  =  35ab.  10.  5mnX9m^ 


MULTIPLICATION.  31 

11.  SaicX— 4^y=  15.    5a"»X— 2a"  = 

12.  —8cmXdn=  16.    3aVX7aV  = 

13.  —7abX2ac=  17.    7aX— 4Z»X— 8c  = 

14.  om^xX3vix^=  18.    SoZ^^  x  3«c  X  —  4^^  = 
^19.    27a^»X  — 39mpXl8ap  = 

20.    6aPy^X2bYX  —  5a^i/  = 
21.7  m^a:;  X  3  mx^  X  —  2  /?iy  = 
22.    —  3pf/  X  6zj^q  X  S2J^(/  = 
23.2  ahu'^x*  X  Sani^x^  X  4  a^mx^  = 

24.  Gx'i/z''  X  —  9x^f/^z^  X  —  3x't/z  = 

25.  3aa;X2amX— 4ma:X^^  = 

26.  7am^  X  3h^n^  X—AahX  aHn  X—2h^n  X  —71171^=^ 

Of  Polynomials  by  Monomials. 

68.    If  we  have  to  multiply  a-\-b  hy  n,  that  is,  to  take 
(a  +  b)  n  times  to  be  added,  we  have, 

(a  +  b)  X  n  =  (a  -\- b) -\- (a -\- b) -{-  (a -\- b) n  times, 

^=  a  -{-  a  -^  (I . . .  71  times  -\-b-{-b-{-b . , . .  n  times, 
=  a  X  ?i-\-b  X  n, 
=  an  -|-  //;/. 

As  it  is  immaterial  in  what  order  the  factors  are  taken, 

n  X  (a  +  b)  =  an  -\-  bn. 
In  like  manner, 

(a-\-b-\-  e)  X  71  =  a7i  -f  bji  -{-  crij 
or,  n  ((I -\- b -\- c)  =  aTi -{-  bn  -\-  en. 


32  ALGEBRA. 

Hence  to  multiply  a  polynomial  by  a  monomial, 

69.  Multiply  each  term  of  the  polynomial  hy  the  monomial^ 
and  add  the  partial  products. 

Exercise  13. 

1.  (6a--5Z>)  X3c  =  18ac  — 15^>c. 

2.  (2  +  3a  — 4a2_5a3)6a2  =  12a2+18a^  — 24a^-30a^ 

3.  ba{U-{-4.G  —  d)  =  15ah-\-2(iac  —  5ad. 

4.  —  ^ax  ( —  hy  —  2cz-\-^')=S  ahxy  •\-  6  acxz  — 15  ax. 

5.  (4a2_3^,)x3aZ»  = 

6.  i^a^  —  <^ab)y.^a^=^ 

7.  {Zx^~\f-\-hz^)'K2x^y=^ 

8.  {a^x  —  5  a^x^  ■\-ax^^2  x')  Xax^y  = 

9.  (—9a'i-3a^'  —  4:aH^  —  b')X—Sab*  = 

10.  (3ic^  — 20^2^  —  7^  +  2/")  X—5:cV  = 

11.  (—4:Xy^  +  5xhj-{-Sx^)X—Sx^y  = 

12.  (— 3  +  2aZ»  +  a2^2>)X  — a^  = 

13.  (—z  —  2xz^-\-5x^yz^  —  6x'f-\-3x^y^z)  X—3xhjz  = 

Of  Polynomials  by  Polynomials. 

70.  If  we  have  a-\-h-\-c  to  be  multiplied  by  m  +  n  +^?, 
we  may  represent  the  multiplicand  a-\-h-{-chj  M.     Then 

M{m  +  ?^  +J9)  =MXm-\- MX  n-^MXp. 

If  now  we  substitute  for  M  its  value, 

(a  •\-  h  -\-  c)(m  -\'  n  -^-p)  ^  (a -\- h -\-  c)  Xm 
+  \a-\-h-^c)Xn 
-\-{a^h-\-c)Xp', 


MULTIPLICATION.  33 

or,  (a  -{-  b  -\-  c)  (m-\-  71  +^)  =  am  -\-  bm  +  cm 

-{•an  -}-b7i  -\-cn 
-\-ap  -^bp  +cp. 
That  is,  to  find  the  product  of  two  polynomials, 

71.  Multiple/  the  multiplicand  by  each  term  of  the  multiplier 
and  add  the  partial  products;  or,  multiply  each  term  of  one 
factor  by  each  term  of  the  other  and  add  the  partial  products. 

72.  In  multiplying  polynomials,  it  is  a  convenient  ar- 
rangement to  write  the  multiplier  under  the  multiplicand, 
and  place  like  terms  of  the  partial  products  in  columns. 

Thus:     (1)    5a  —    6  6 

3a  —   4  6 

\ha^  —  \^ah 

—  20a6  +  246' 
15a2-38a6-|-246* 

(2)  Multiply  4x4-3  +  5x2  — 6a:«  by  4  —  6x2  — 5ic. 
Arrange  both  multiplicand  and  multiplier  according  to 

the  ascending  powers  of  x. 

3+   4x+    5x2—    g^ 
4 —   5x —   6x2 
12-hl6x  +  20x2-24x« 

-15x-20x2  — 25x«  +  30x* 

-18x2  — 24x^  — 30x*  +  36x» 

12+      X— 18x2  — 73x»  +36x» 

(3)  Multiply  l  +  2x  +  x*  — 3x2  by  x3  — 2  — 2x. 
Arrange  according  to  the  descending  powers  of  x. 

x*-3x2  +  2x  +1 

or^  — 2x  — 2 

x'^-3x^  +  2x*+    x« 

—  2x*  +6x^  —  4x2  — 2x 

— 2x^  +6x2  — 4.r  — 2 

a^  — 5x«  +7x«  +  2x2  — 6x  — 2 


34 


ALGEBRA. 


(4)    Multiply  a^'\'b^-^c^  —  ah  — bo  — achy  a  +  h-^c. 
Arrange  according  to  the  descending  powers  of  a, 

€?  —  ah  —  ac-\-    h^  —      Z>c  +    (? 

a  -\-     h-\-     c 

a^  —  a%  —  a^c  +  aV^  —    ahc  +  (t^ 

'\-a%  —ah^~    ahc  -\- h^  —  h^c -\- hc^ 

-\-a?G  —     ahc  —  ac^  -\-h'c  —  hc^-\-G^ 


a^  —'dahc  -\-ly'  -\-(^ 

The  student  should  observe  that,  with  a  view  to  bringing 
like  terms  of  the  partial  products  in  columns,  the  terms  of 
the  multiplicand  and  multiplier  are  arranged  in  the  same 
order. 

In  order  to  test  the  accuracy  of  the  work,  interchange 
the  multiplicand  and  multiplier.  The  result  should  be  the 
same  in  both  operations. 


Exercise  14. 
Multiply : 

1.  rr^  — 4by  a;2+5.  3.  a^  +  a V  +  a^*  by  a^ _ ^2. 

2.  y  —  6  by?/ +  13.  4.    x^-^xy-^-if  hj  x  — y. 

5.  2x  —  yhj  x-\-2y. 

6.  2^^  +  4a^2_^8a;  +  16by  3ic  — 6. 

7.    x^-\-x^-^x  —  1  by  X  —  1.        8.    x^  —  ^ax\)jx-\-^a. 
9.    2h^^3ab  —  a^hj  —Bh-{-la. 

10.  2a-\-hhj  a-\-2h.  12.    a^— ah-^h^hj  a-\-h. 

11.  a^-\-ab-{-h^hy  a  —  h.  13.    2 ah  — oh"^  hy  3 a^—4:ab, 

14.  —a^-[-2a''h'-h^hy^a^  +  ^ah, 

15.  a'''{-ah-\'h''hy  a^  —  ah-\-h\ 


MULTIPLICATION.  35 

16.  a^  —  SaH  +  3aP  —  b''hja^  —  2ab-^l^, 

17.  x-\-22/  —  3zhy  x  —  2y-]-3z. 

18.  fic2  +  3ic?/  +  42^by  3x2  — 4a'2/  +  y«. 

19.  x^-\'X//-{-i/^hy  ay^-{-xz-\-z^. 

20.  a^  +  //  +  6-2  —  ab  —  ac  —  bc  by  a  +  />  +  c. 

21.  a,'2  — j-y-f-?/2_^.x+7/  +  l  by  ic  +  7/  — 1. 

Arrange  the  multiplicand  and  multiplier  according  to 
the  descending  powers  of  a  common  letter,  and  multiply : 

22.  5x-\-4:X^  +  x^  —  24:hyx^-{-ll—4x.    — 

23.  a-'4-lla-  —  4.T2-- 24  by  .7-2  +  5  +  40!. 

24.  x'-\-x^~4x  —  ll-\-2xn^yx^  —  2x-{-3.    — 

25.  —oj-*  —  x-  —  x  +  x'-\-13x^hyx^  —  2  —  2x, 

26.  3ir  +  a;3  — 2:r2  — 4by  2ic+4ic«  +  3a:2+l. 

27.  5 a*  +  2 a^^b^ -\-ab^  —  3w'bhy  5 ci'b  —  2 ab""  +  3 a%^  +  b\  -^ 

28.  4  a^  —  32  ai/  —  8  aV  + 16  ay  by  a V  +  ^  ^^V  +  4  ay. 

29.  3m3  +  3?i«  +  9?/i7i2  +  97/i2^by  6?7iV  — 2wm* 

30.  Qa''b-{-3a^b'~2(ib'-{-b''  by  4a*-2aZ»"  — 36^ 
Find  the  products  of : 

31.  a;  — 3j  X  — 1,  a-  +  l,  and  ic  +  3. 

32.  ic-  — x+1,  a;2  +  .r  +  l,  andic^  — x^  +  l. 

33.  tr"  +  ab  +  />^  a^  —  ab-\-  b%  and  a^  -  a^b^  +  Z»^ 

34.  4a3-4a2^,  +  «^2,  4.a^-^3ab  +  b\  and  2a2Z,  +  ^,3, 

35.  £c  +  a,  cc  +  2a,  cc  —  3  a,  a:;  —  4  a,  and  £c  +  5  a. 
36.9  a^  +  /y2,  27  a^  —  ^»3,  27  a^  +  b\  and  81  a*  —  9  a^Z.^  _|.  j4^ 


36  ALGEBRA. 

37.  From  the  product  oiif  —  ^yz^    z^  and  ?/^  +  2 y^  —    z^ 

take  the  product  of  y^  —    yz  —  2z^  Siud  y^-\-    yz  —  2  z\ 

38.  Find  the  dividend  when  the  divisor  =Sa^  —  ab  —  3b% 

the    quotient   =a^b  —  2b^  the    remainder   =  —  2ab^ 
—  6b\ 

The  multiplication  of  polynomials  may  be  indicated  by 
inclosing  each  in  a  parenthesis  and  writing  them  one  after 
the  other.  When  the  operations  indicated  are  actually  per- 
formed, the  expression  is  said  to  be  simplified. 

I    Simplify : 
^9.    (a-\-b~c)(a-\-c  —  b)(b-\-c  —  a)(a'\-b-\-c), 
\40.    (a  +  b){b-\-c)  —  (c-\-d)(d-\-a)--{a-^c){b  —  dy 

41.  {a-\-b  +  c-\-dy-\-{a  —  b  —  c-\-dy 

-\-{a  —  b  +  c  —  dy-\-{a-{-b  —  c  —  dy. 

42.  (a-\-b-\'cY—a{b-\-G  —  a)  —  b{a  +  c~b)  —  c{a-{-b  —  c). 

43.  {a--b)x  —  (b--c)a--\{b'-x){b-~a)  —  {b  —  c){b-\-c)}. 

44.  (m -}- 7i)m  —  J  (m  —  ny  —  (n  —  7ri)n\. 

45.  {a-b-\'cy—\a{G  —  a  —  b)  —  [b{a-\'b  +  G) 

—  c(a  —  b  —  c)']\, 

46.  {p'-^qy'-'{p'^q){p\r-'q\-q\r'-2>\y 

47.  {^xhf'-4.y'')(x'  —  if)—\3xy''2y^\ySx{x'-\-f) 

-'2y{f'\-3xy-x^\y, 

48.  a^—\2ab'-[;-(a-{-\b--c\)(a  —  \b'-c\)'\-2ab'] 

-Ucl  —  qj-^-cf. 

'49.    \ac—{a  —  b){b-\-c)\  —  b\b  —  {a-c)\. 

50.  5|(a  —  b)x  —  cy\  —  2\a(x  —  y) — bx\ 

—  \3ax  —  {nc^2a)y\, 

51.  (a;-l)(a;  — 2)-3a;(a;  +  3)+2J(ic  +  2)(a;-i-l)— 3^. 


MULTIPLICATION.  37 


52.  l(2a-]-by-j-(a  —  2byiX\(Sa  —  2by—(2a  —  3by\. 

53.  4.(a-3bXa-\-3b)-2(a-6by-2(a^-i-6b^). 

54.  x\x'  +  i/y-2xy(x-{-yXx-y)-(x'-f)\ 

55.  16C«2  +  Z,2^(a2-^»^)-(26i-3)(2a  +  3)(4a2  +  9) 

-\-{2b-3)(2b  +  SXW-{-9y 


73.  There  are  some  examples  in  multiplication  which 
occur  so  often  in  algebraical  operations  that  they  should 
be  carefully  noticed  and  remembered.  The  three  which 
follow  are  of  great  importance  : 


(1)    a  +      b 

(2) 

a  -       b 

(3) 

a  -\-   b 

a  -\-      b 

a  —      b 

a  -    b 

a^-\-    ab 

a^—    ab 

a^-\-ab 

ab  +  b' 

-    ab-\-b^ 

—  ab-b^ 

a^-{-2ab-\-b' 

a^-2ab-^b'' 

a^          -b^ 

From  (1)  we  have  {a  +  bf  =  a^-\-2ab-\-  b\     That  is, 

74.  The  square  of  the  sum  of  two  numbers  is  equal  to  the 
sum  of  their  squares  -\-  twice  their  product. 

From  (2)  we  have  (a  —  bf  =  a^  —  2ab  +  b\     That  is, 

75.  The  square  of  the  difference  of  two  numbers  is  equal 
to  the  sum  of  their  squares  —  twice  their  product. 

From  (3)  we  have  (a  -{■b){a  —  b)  =  a^  —  b\     That  is, 

76.  The  product  of  the  sum  and  difference  of  two  numbers 
is  equal  to  the  difference  of  their  squares. 

11.   A  general  truth  expressed  by  symbols  is  called  a 
formula. 


38  ALGEBRA. 

78.  By  using  the  double  sign  ±,  read  plus  or  minus,  we 
may  represent  (1)  and  (2)  by  a  single  formula ;  thus, 

in  which  expression  the  upper  signs  correspond  with  each 
other,  and  the  lower  signs  with  each  other. 

By  remembering  these  formulas  the  square  of  any  bino- 
mial, or  the  product  of  the  sum  and  difference  of  any  two 
numbers,  may  be  written  by  inspection  ;  thus  : 


Exercise  15. 

1.  (127)2-  (123)2=  (127  + 123)  (127  - 123) 

=  250X4  =  1000. 

2.  (29)2=  (30 -1)2  =  900 -60 +  1  =  841. 

3.  (53)2  =  (50  _^  3)2  ^  2500  -f  300  +  9  =  2809. 

4.  (3cc  +  22/)2  =  9cc2  +  12cc?/  +  4?/2^ 

5.  {2a'x  —  bx'ijf  =  4.a^x'  —  20a'^x^y-\-2^x^y\ 

6.  (3a^»2c-f-2a2c2)(3a^»2c_2aV)  =  9a2Z;4c2  — 4a^c*. 

7.  (x-\-yy=  15.    (ah-\-cdf  = 

8.  {y  —  zy=  16.    (3m?^  — 4)2  = 

9.  (2a: -4-1)2=  17^    (12  +  5ic)2  = 

10.  (2a-\-Uf=  18.    (4:X2f-yzy  = 

11.  (1  — a;2)2=  19.    {3ahG  —  hcdf  = 

12.  {^ax  —  ^xy=        20.    (4.x^  —  xi/)''  = 


13.  (1  — 7a)2=  21.    {x-{-y){x  —  ij)  = 

14.  {bxy^2y=  22.    {2a-\-h)(2a-b)  = 


MULTIPLICATION.  o9 

23.  (S—x)(S-\-x)  = 

24.  (3ab-\-2b^)(3ab  —  2b^  = 

25.  (4.x^-Sf)(Ax^-^Sf)  = 

26.  (a3«2  —  by*)  (a^x^  +  ^>?/^)  = 

27.  (6a:2/-5y2)(6a:?/  +  5y2)  = 

28.  (4x«-l)(4ir«  +  l)  = 

29.  (l-\-3ab^)(l-3ab^)  = 

30.  (ax  +  by)  (ax  —  by)  (a^a^  +  b^f)  = 

79.    Also  the  square  of  a  trinomial  should  be  cai*efully 
noticed. 

a  -{-  b-\-  c 
a  -\-  b-\-  c 
a^-\-    cib-{-    ac 

ab  -\-b^+    be 

ac         -{-    bc-\-c^ 

a^-\-2ab-\-2ac-}-b^  +  2bc-{-c^, 

=  a^-\-b'-\-c^-\-2ab  +  2ac  +  2bc. 

It  IS  evident  that  this  result  is  composed  of  two  sets  of 
numbers : 

I.    The  squares  of  a,  b,  and  c ; 
II.    Twice  the  products  of  a,  b,  and  c  taken  two  and  two. 

Again, 


a  — 

b- 

c 

a  — 

b- 

c 

a'- 

ab  — 

ac 

— 

ab 

+  ^»2-f.      be 

— 

ac         -\-    bc-\-  c* 

a^-'2ab  —  2ac-^b^-{-2bG-\-c^ 
■  a^  +  b^  +  c^  —  2ab  —  2ac-}'2bc. 


40  ALGEBRA. 

The  law  of  formation  is  the  same  as  before : 

I.    The  squares  of  a,  b,  and  c; 
II.    Twice  the  products  of  a,  b,  and  c  taken  two  and  two. 

The  sign  of  each  double  product  is  +  or  —  according  as 
the  signs  of  the  factors  composing  it  are  like  or  unlike. 

The  same  law  holds  good  for  the  square  of  expressions 
containing  more  than  three  terms,  and  may  be  stated  thus  : 

80.  To  the  sum  of  the  squares  of  the  several  terms  add  twice 
the  product  of  each  term,  by  each  of  the  terms  that  follow  it. 

By  remembering  this  formula,  the  square  of  any  poly- 
nomial may  be  written  by  inspection ;  thus  : 

Exercise  16. 

1.  (x-\-^j-\-zy=  9.  (a'^  +  ^,3-f  c3)2  = 

2.  (x  —  y-{-zy=  10.  (x^  —  y^  —  zy  = 

3.  {7ti-\-n—p  —  qy=  11.  (cc  +  2?/  — 3.^)2  = 

4.  {x^-\-2x  —  ^f=  12.  (x^  —  2y^^bzy  = 

5.  {x^  —  &x-\-lf=  13.  {x^-\-2x-~2y  = 

6.  {2x''—lx-\-^y=  14.  (x^  —  bx-{-iy  = 

7.  (x'-\-y''  —  z:'y=  15.  {2x^  —  ^x  —  ^y  = 

8.  (a;^  — 4a;2/  +  y')'=  16.  (.t  +  2?/  +  3^)2  = 

81.  Likewise,  the  product  of  two  binomials  of  the  form' 
x-\-a,  x-l-b  should  be  carefully  noticed  and  remembered. 


(1)   0^+5 

(2)    x-5 

X  +3 

X  -3 

x^-\-^x 

x^^^x 

+  3x4-15 

-3:r  +  15 

0^2+ 8a; +  15 

a;2-8a;  +  15 

MULTIPLICATION.  41 


(3)    X  +5  (4)    X  -5 

X  —3  ic  +3 


7?-\-bx  x^  —  6x 

—  3a;  — 15  +3^;- 15 

a:^  +  2a,--15  a^-2a;-15 


It  will  be  observed  that : 

I.  In  all  the  results  the  first  term  is  x^  and  the  last  term 
is  the  product  of  5  and  3. 

II.  From  (1)  and  (2),  when  the  second  terms  of  the 
binomials  have  like  signs,  the  product  has 

the  last  term  positive; 

the  coefficient  of  the  middle  term  =  the  sum  of  3  and  5 ; 
the  sign  of  the  middle  term  is  the  same  as  that  of  the 
3  and  5. 

III.  From  (3)  and  (4),  when  the  second  terms  of  the 
binomials  have  unlike  signs,  the  product  has 

the  last  term  negative; 

the  coefficient  of  the  middle  term  =  the  difierence  of 
3  and  5; 

the  sign  of  the  middle  term  is  that  of  the  greater  of  the 
two  numbers. 

82.  These  results  may  be  deduced  from  the  general 
formula, 

(x-\-a)(x-{-h)=x''^{a-\-b)x-\-ab, 

by  supposing  for  (1)  a  and  b  both  positive; 

(2)  a  and  b  both  negative; 

(3)  a  positive,  b  negative,  and  a  >  ^ ; 

(4)  a  negative,  b  positive,  and  a'>b. 

By  remembering  this  formula  the  product  of  two  bino- 
mials may  be  written  by  inspection ;  thus  : 


42  ALGEBRA. 


Exercise  17 

. 

1. 

(x-\-2)(x^^)  = 

11. 

{x  —  G)(x  —  cT)  = 

2. 

(a;  +  l)(x  +  5)  = 

12. 

(x  —  4.y){x-^y)  = 

3. 

{x-^)(x-Q>)  = 

13. 

(a  —  2b)(a  —  5b)  = 

4. 

{x-^)(x-l)  = 

14. 

(x'-\-2f)(x^  +  f)  = 

5. 

{x-^){x  +  l)=^ 

15. 

(x'-Sxy)(a^-{-xy) 

6. 

(^-2)(^  +  5)  = 

16. 

(ax  —  9)(ax  +  6)  = 

7. 

(x-^){x-^l)  = 

17. 

(x-]-a)(x-b)  = 

8. 

(x-2)(x-4.)  = 

18. 

(x-ll)(x  +  A)  = 

9. 

(x^l)(x-^lV)  = 

19. 

(a. +  12)  (0.-11)  = 

10. 

(x  —  2a){x-^^ay. 

=                 20. 

(^-10)(^-5)  = 

83.    The  second,  third,  and  fourth  powers  oi  a-{-b  are 
found  in  the  following  manner  : 


a  -\-      b 

a  -{-      b 

a^-{-    ab 

ab-\-P 

(a  +  by-- 

=  a'-{-2ab-{-P 

a  -i-b 

a^-\-2a'b-}-    ab' 

a'b-{-2aP-}-b^ 

(a  +  by-- 

=  a'-{-3a'b-i-Sab'-{-b^ 

a  -\-b 

a*i-3a%-\-3a%'+    ab^ 

a%-{-3a%'-{-Sab^-{-b'' 

(a-{-by  =  a*-{-4:a^b-{-6a^'-}-4:ab^-}-b'' 


MULTIPLICATION.  43 

From  these  results  it  will  be  observed  that : 

I.  The  number  of  terms  is  greater  by  one  than  the  ex- 
ponent of  the  power  to  which  the  binomial  is  raised. 

II.  In  the  first  term,  the  exponent  of  a  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised; 
and  it  decreases  by  one  in  each  succeeding  term. 

III.  h  appears  in  the  second  term  with  one  for  an  expo- 
nent, and  its  exponent  increases  by  one  in  each  succeeding 
term. 

IV.  The  coefficient  of  the  first  term  is  1. 

V.    The  coefficient  of  the  second  term  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised. 

VI.  The  coefficient  of  each  succeeding  term  is  found 
from  the  next  preceding  term  by  multiplying  its  coefficient 
by  the  exponent  of  a,  and  dividing  the  product  by  a  num- 
ber greater  by  one  than  the  exponent  of  b. 

84.  If  b  is  negative,  the  terms  in  which  the  odd  powers 
of  b  occur  are  negative.     Thus  : 

(a  —  by  =  a'-4.a^b^e>a^b''  —  4.ab^^b\ 


Exercise  18. 

Write  by  inspection  the  results  : 

1.  {x-\-aY=                  5.    (x^aY=  9.  (x  +  y)*  = 

2.  (x  —  (if=                 6.    {x  —  ay=  10.  (x  —  yy  — 

3.  (cc+i)«=           7.  (x-\-iy=  11.  (x-^iy= 

4.    {x  —  iy=                  8.    (a^  — 1)^=  12.  {x  —  iy== 


CHAPTER   IV. 

Division. 

86.  Division  is  the  operation  by  whicli,  when  a  product 
and  one  of  its  factors  are  given,  the  other  factor  is  deter- 
mined. 

86.  With  reference  to  this  operation  the  product  is  called 
the  dividend ;  the  given  factor  the  divisor ;  and  the  required 
factor  the  quotient. 

87.  The  operation  of  division  is  indicated  by  the  sign  -i- ; 

by  the  colon  :,  or  by  writing  the  dividend  over  the  divisor 

12 
with  a  line  drawn  between  them.    Thus,  12  -f-  4,  12  :  4,  —, 

each  means  that  12  is  to  be  divided  by  4. 

88.  + 12  divided  by  +  4  gives  the  quotient  +  ^ ;  since 
only  a  positive  number,  +3,  when  multiplied  by  +4,  can 
give  the  positive  product,  + 12.  §  61. 

+ 12  divided  by  —  4  gives  the  quotient  —  3  ;  since  only 
a  negative  number,  —  3,  when  multiplied  by  —  4,  can  give 
the  positive  product,  -j- 12.  §  61. 

— 12  divided  by  -j-  4  gives  the  quotient  —  3 ;  since  only 
a  negative  num-ber,  —  3,  when  multiplied  by  +  4,  can  give 
the  negative  product,  — 12.  §  61. 

— 12  divided  by  — 4  gives  the  quotient  +3  ;  since  only 
a  positive  number,  +  3,  when  multiplied  by  —  4,  can  give 
the  negative  product,  —12.  §  61. 


DIVISION.  45 

(1)  ^  =  +  3-  (3)   ^  =  -3. 

(2)  ^  =  -3-  (4)    ^  =  +  3. 

From  (1)  and  (4)  it  follows  that 

89.  The  quotient   is  positive  when  the   dividend   and 
divisor  have  like  signs. 

From  (2)  and  (3)  it  follows  that 

The  quotient  is  negative  when  the  dividend  and  divisor 
have  unlike  signs. 

90.  The  absolute  value  of  the  quotient  is  equal  to  the 
quotient  of  the  absolute  values  of  the  dividend  and  divisor. 


Exercise  19. 


+  264           +3840  106.33 

1.    .  .  =         3. r:— =         5. 


—  30  —4.9 


-2568  -42.435 

2. 4.  --—-—=        6. 


+  4 

—                    « 

-4648  = 

-8 

7. 

-264 
+  24 

Q 

-3670 

—  85 

9. 

+  6.8503 

+  12  +34.5 

-7.1560 
10. 


+  324 

—  1 
—  3.14159 

-0.31831 
—  61  ~         *"•  —31.4159 


11. 


12. 


46  ALGEBRA. 

Division  of  Monomials. 

91.  If  we  have  to  divide  abc  by  he,  aahx  by  aby,  12abo 
by  —  4  ab,  we  write  them  as  follows  : 

abc  aabx      ax  12  abc 

~--  =  a.  —7-= — •  — T— 7=  — dc. 

be  aby        y  — 4a6 

Hence,  to  divide  one  monomial  by  another, 

92.  Write  the  dividend  over  the  divisor  with  a  line  between 
them;  if  the  expressions  have  common  factor s^  remove  the 
common  factors. 

If  we  have  to  divide  a^  by  a^,  a^  by  a^,  a^  by  a,  we  write 
them  as  follows : 


a-" 

aaaaa 

-aaa 

=  a\ 

a'' 

aa 

a' 

aaaaaa 

=  a\ 

a' 

aaaa 

«'_ 

aaaa 

aaa- 

=  a?. 

93.  That  is,  if  a  power  of  a  number  is  divided  by  a 
lower  power  of  the  same  number,  the  quotient  is  that  power 
of  the  number  whose  exponent  is  equal  to  the  exponent  of  the 
dividend  minus  that  of  the  divisor. 

Again, 


0?         aa            11 

a^      aaaaa      aaa      a? 

€?        aaa         1        1 

a^      aaaaa      aa      a^ 

a*           aaaa              1 

1 

a^      aaaaaaaa      aaaa 

a' 

DIVISION.  47 

94.  That  is,  if  any  power  of  a  number  is  divided  by  a 
higher  power  of  the  same  number,  the  quotient  is  exjjressed 
by  1  divided  by  the  number  with  an  exponent  equal  to  the 
exponent  of  the  divisor  minus  that  of  the  dividend. 

Exercise  20. 
-\-ah ,  10  0^^ —Sbmx 


-\-a                          '     2bG  '       4:ax^ 

4-ab           ,                     x^  ab^(? 
=  —  b.  8.    ^=  14.    — -  = 

—  a                              —  cr  abc 

—  ab           ,                   —12  am  m^p^x^ 
=  —  b.  9. =  15.       ^ 


+  a           ""  *"      — 27?i  *"*    mp^x^ 

—  ab       .   .  35  abed                         — 51  abdy^ 
=~rb.  10.                 =  16.    — ^73 = 

—  a  5bd                                Sbdy 

6mx abx  225 m^y 

2x  *    5aby                        '    25 my^ 

12  a*  „      27  a^  _     SOo^y 

— — =  12.    — -—.=  18.    — ^-|-= 

—  3a  —  3a^  •        —5xy 

4:a''m*x'  —3a%h*d'' 


19.    -— — ^=  21. 


5  a^//i^a;  '      —  a%^cd 


^2x?fz*_  12  amHY(f  _ 

^^*      lx.f^~  ^^'      4mVi>V   ~ 

23.  (4 a%^  X  10  a2^,3;^)  -v-  5  a^^^;^^  = 

24.  (21  a^VV  -J-  3  irv/2.^)  (-  2  .ry;^)  = 

25.  104  a5V  -^  (91  «V/x^  -f-  7  aV/o-)  = 

26.  (24  aWx  -f-  3  a^^^s)  +  (35  a%''x'  -i-  —  5  a%x)  = 

27.  85a*^+i^5a*'"-2=  28.    84  a"-" -^  12  «» = 


48  ALGEBRA. 

Of  Polynomials  by  Monomials. 

96.    The  product  oi  (a -\- b -{- c)  X2^  =  ccp-{-  h^  +  cp. 
If  the  product  of  two  factors  is  divided  by  one  of  the 
factors,  the  quotient  is  the  other  factor.     Therefore, 

(ap  -\rhp-\-  cp)  -^  p^=^a-\-h-\rC. 

But  a,  i,  and  c  are  the  quotients  obtained  by  dividing 
each  term,  ap^  hp,  and  cp  by  p. 

Therefore,  to  divide  a  polynomial  by  a  monomial, 

96.    Divide  each  term  of  the  polynomial  hy  the  monomial. 


Exercise  21. 

1.  (Sab  —  12ac)^4.a  =  2b  —  3c. 

2.  (15am  —  10bm  +  20cm,)^—5m  =  —  3a-^2b  —  4:c. 

3.  (ISamy  — 27 bny-}- 36 cpy)~—9y  = 

4.  (21ax  —  lSbx-{-15cx)^  —  3x  = 

5.  (12x'  —  Sx^  +  4:x)^4.x  = 

6.  (3x^  —  6x''-\-9x'  —  12x^)-^3x^=^ 

7 .  (35  m^y  -f-  28  my  — 14  7ny^)  ~-  —  7  my  = 

8.  (4  a'^b  —  6  a%^  + 12  a^b^)  -^2a%  = 

9.  (12 xhf  — 15 xY  —  24 x'y)  -^—3x^y  = 

10.  (12a:y-24a;y +  36^y-12a;y)  -V-  12a^y  = 

11.  (3a''  —  2a'b-a%^)-i-a''  = 

12.  (3xhjz^-{-6x^yz^  —  15xYz^-\-lSxyz)  -i-—3x^yz  = 

13.  (— 16  a%\^  -f  8  a^bh'  — 12  a'bh^)  -f-  —  4  a%h^  = 


division.  49 

Of  Polynomials  by  Polynomials. 

97.  If  the  divisor  (one  factor)  =  a-\-b-{-c, 
and  the  quotient  (other  factor)  =              ^i+i?  +  2'j 

r      a7i  -\-bn-{-  en 
then  the  dividend  (product)        =  J  -\-  ap-^-hjp-^-cp 

The  first  term  of  the  dividend  is  an  j  that  is,  the  product 
of  a,  the  first  term  of  the  divisor,  by  ti,  the  first  term  of  the 
quotient.  The  first  term  n  of  the  quotient  is  therefore 
found  by  dividing  an^  the  first  term  of  the  dividend,  by  a, 
the  first  term  of  the  divisor. 

If  the  partial  product  formed  by  multiplying  the  entire 
divisor  by  n  is  subtracted  from  the  dividend,  the  first  term 
of  the  remainder  ap  is  the  product  of  a,  the  first  term  of 
the  divisor,  by  ^,  the  second  term  of  the  quotient.  That  is, 
the  second  term  of  the  quotient  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  first  term  of  the  divisor. 
In  like  manner,  the  third  term  of  the  quotient  is  obtained 
by  dividing  the  first  term  of  the  new  remainder  by  the  first 
term  of  the  divisor,  and  so  on. 

Therefore,  to  divide  one  polynomial  by  another, 

98.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor. 

Write  the  result  as  the  first  term  of  the  quotient. 

Multiply  all  the  terms  of  the  divisor  by  the  first  term  of 
the  quotient. 

Subtract  the  product  from  the  dividend. 

If  there  is  a  remainder,  consider  it  as  a  new  dividend  and 
proceed  as  before. 


50  jIlgebra. 

99.  It  is  of  great  importance  to  arrange  both  dividend 
and  divisor  according  to  the  ascending  or  descending  powers 
of  some  common  letter,  and  to  keep  this  order  throughout  the 
operation. 


Divide 

Exercise  22. 

) 

(1)    a^-^2ab  +  b'] 

by  a  +  J ; 

(2) 

a^  —  b^hj  a-\-b', 

a'-^2ab-]-b''[ 

^2+    ab 

ab-^b^ 
ab-\-b^ 

a-\-b 
w^b 

a^-    b''\a-^b 
cu^-\-ab   a  —  b 

—  ab  —  b^ 

—  ab  —  b^ 

(3) 

a'-2ab-{-b''hj 

a  —  b 

i 

a^-2ab-\-b^\a- 
c?  —    ab           a  — 

-b 
-b 

—  ab-{-b^ 

—  ab-i-P 

(4)    4a^ic 

:2  — 4aV  +  aj^  — a« 

by  x^ 

-a^', 

x'- 
x'- 

a'x' 

x'- 

a" 
3aV+a* 

— 

36^V  +  4aV-a« 

- 

a'^x^  —  a^ 
a^x^  —  a^ 

(5)    22a%^-^lDb^-\-Za^-l(^a^b-22ab^hY  a^-\-Sb''-2ab', 
3 ^4 _  10 a^b  +  22 a^b^ - 22 ab^  + 15 b'\    a^-2ab-\-3b^ 
3a'-   ea'b+   9a'P  3a^  —  4:ab-\-5P 

—  Aa^b -}- IS  a%^  — 22  ab^ 

—  4:a^b+   Sa^b^  —  12ab^ 


5a^b^-10ab^-\-15b' 
5  a^b^- 10  ab'  + 15b* 


DIVISION.  61 


Divide 


'  6.  x^  —  7x-}-12hj  x  —  3. 

7.  ic^  +  ic  — 72  by  a;  +  9. 

8.  2x^  —  x''  +  Sx  —  9hj2x  —  3. 

9.  6x'  +  Ux^  —  Ax  +  24:hy2x  +  6. 

10.  3a;24-x  +  9ic3  — 1  by  3a;  — 1. 

11.  7x'-\-5Sx  —  24:X^  —  21\)j7x  —  3. 

12.  aj^  — 1  by  x  — 1. 

13.  a^  — 2o^»2_|.^8by  «_j. 

14.  a;*  — 81 2/*  by  x  — 3  y. 

15.  rr^  —  y  by  ic  —  y. 

16.  a^  +  32Z'«by  a  +  2^i. 

17.  2a*-{-27aP  —  SU*hj  a-{-Sb. 

18.  a;^4-ll^'-12a^-5a:«+6by  3  +  x2_3^^ 

19.  a;^  — 9x2  +  ^3  — 16x  — 4by  x2  +  4-f-4ar. 

20.  36+a:^  — 13ic2|3y  g_p3,2_^5^ 

21.  a;*  +  64by  icH4a;  +  8. 

22.  x*  +  x^-}-57  —  S5x  —  24.x^hjx'-3-^2x. 

23.  1-0^  —  3x2  — x«  by  l  +  2a;  +  x2. 

24.  x^  —  2x^  +  lhj  x^  —  2x-{-l. 

25.  a^  +  2rt2^''^  +  96*by  a2-2aZ'  +  3^''. 

26.  4a;^  — x^  +  4icby  2  +  2x2  4-3a. 

27.  a*  — 243  by  a  — 3. 

28.  18a:*  +  82a;*+40  — 67x  — 45a^by  3a:*  +  5  — 4x. 

29.  x*  —  6xj/—9x^'-fhyx^-\-y-\-3x. 


52  ALGEBHA. 

30.  x^-\-^oi?y^  —  Qx^y  —  4.y^\)yx'—^xy  +  2f. 

31.  x^-\-x^y^-^y^hy  x^  —  xy-\-y^. 

32.  x'^-\-x^-\- xSj  +  ]f'  —  2xy^  —  xV  by  x''  +  £c  —  y. 

33.  2x^  —  ^'if-\-xy  —  xz'-^yz  —  z^\^y  2x-{'3y  +  z. 

34.  12  +  82a;2  +  106a;*-70:c^-112a:3__3g^ 

hj  S-Bx-i-Tx^ 

35.  33*  +  ?/^  by  a;*  —  x^^  +  a^^y'^  —  xy^  -\-  y*. 

36.  2a:^  +  2«y  —  2xf—  Ix^y  —  i/ hy  2 x"^ -{- ^f  —  xy. 

37.  16ic^  +  4xy  +  ?/^by  4x2  — 2iC2/  +  2/l 

38.  32a«^>  +  8a3^'«  — a^»*  — 4a'^i*  — SGa^^'^ 

by  ^>3_4^2j^6^j2^ 

39.  1  +  5x3  — 6x*  by  l—x  +  3x2. 

40.  1  —  52  a^^.*  —  51  a%^  by  4  a%^  +  Zab  —  1. 

41.  x^?/  —  x^  by  x^^  +  2  xif  —  2  x^  —  y^. 

42.  x6+15xy+15xy  +  i/«-6x«2/  — 6xz/*  — 20xy 

by  x^  — 3x2z/  +  3xy2  — 2/8. 

43.  aJ-\-2a%^  —  2a^b^-2a^h  —  Qa%'^  —  3ab^ 

hy  a^  —  2a%  —  ah\ 

44.  81xV  +  18xy  —  54xy  —  18xy  —  18x/—  9/ 

by  3x*  +  xy  +  2/' 

45.  a*  +  2(i«Z»  +  8a2^»24.8aJ3+16^>^by  a2  +  4^»*. 

46.  8^^- x«+21xy  — 24x/by  3x3/  — x'^  — 2/2. 

47.  16a*  +  9J^  +  8a2j2by4^2_|_3j2_4^j^ 

48.  as_|.j3^^s_3^j^l3y  ^_^j_j_c. 

49.  a^-{-W^c^  —  Qahc\iy  a^  +  W^ -[■  c^  —  ac  —  2 ah  —  2hc. 

50.  «8  +  ^»3_|.^3^3^2^,_|.3^^,2l3y  ^_^j_^^. 


DIVISION.  53 

100.    The  operation  of  division  may  be  shortened  in  some 
cases  by  the  use  of  parentheses.     Thus : 

a^ -jr  (a -\- b -\- c)  x^ -\-  (ah -\- ac -\- be) x -\-  abc \x  -\-b 


a^'+(   +^        )x'' x''-\-{a^c)x-\-ac 

(a         -]-c)x^-\-  (ab  -\-ac-\-  be)  x 

(a        -\-e)x^-{-  (ab -\-bc)x 

acx  +  «^^' 

dcx  +  abc 


Exercise  23. 
Divide 

1.  a^(b  +  c)-}-b^(a  —  c)-\-c\a  —  b)  +  abchy  a-\-b  +  c. 

2.  x^  —  (a -j- b -{- c)  x^ -\-  (ab  +  ^*c  -\-bc)x  —  abc 

by  a;'^  —  (a-{-b)x-{-  ab. 

3.  x^  —  2ax''+(a^-^ab  —  b^x  —  a%  +  aPhyx  —  a-{-b. 

4.  x*  —  (a^  —  b  —  c)x^  —  (b  —  e)ax-{-bc  by  x^  —  ax-\-  c. 

5.  ]^  —  (iR  -\-  n  +i-?)y^  "I"  (^'i'i  4"  ^M^  ~l~  ''^p)y  —  'mnp  by  y  — p. 

6.  a;*+(5  +  a)a:3— (4  — 5a  +  ^»)a:2— (4a  +  5Z')£c  +  4Z» 

by  x^-^^x  —  4. 

7.  x^—(a^b^c-{-d)x''^(ab-{-ac^ad-\-bc-\-bd-^ed)x'' 

—  (abc  -|-  aJ)d  -\-  acd + bcd)x  -\-  abed 

by  x^  —  (^  +  o)x  -\-  ac. 

8.  oj*  —  (m  —  c)x^-\-  (n  —  em  -\- d)  x^ -\-  (r -{-  eri  —  dm)  x^ 

-\-  (cr  -\-  dn)  x-\-dr\ij  7^  —  mx^  -{-nx-^r. 

9.  a"  —  mx^  -\-  nx^  —  nx^-\-mx  —  1  by  a;  —  1. 

10.    (x-\-yy^3(x-^yyz^-^(x-\-y)z'^z' 

hy  (x-]-yy^2(x-^y)z-\-z\ 


54  ALGEBRA. 

101.  There  are  some  cases  in  Division  which  occur  so 
often  in  algebraic  operations  that  they  should  be  carefully 
noticed  and  remembered. 


Case  I. 
The  student  may  easily  verify  the  following  results : 

^  ^     a  —  b 
^  ^     a  —  0 

^5 OO  75 

^  ^      a  —  2o 

From  these  results  it  may  be  assumed  that : 

102.    The  difference  of  two  equal  odd  powers  of  any  two 
numbers  is  divisible  by  the  difference  of  the  numbers. 


It  will  also  be  seen  that : 

I.  The  number  of  terms  in  the  quotient  is  equal  to  the 
exponent  of  the  powers. 

II.  The  signs  of  the  quotient  are  all  positive. 

III.  The  first  term  of  the  quotient  is  obtained,  as  usual, 
by  dividing  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor. 

ly.  Each  succeeding  term  of  the  quotient  may  be  ob- 
tained by  dividing  the  preceding  term  of  the  quotient  by 
the  first  term  of  the  divisor,  and  multiplying  the  result  by 
the  second  term  of  the  divisor  (disregarding  the  sign). 


division.  £5 

Exercise  24. 

Write  by  inspection  the  results  in  the  following  exam- 
ples: 

1.  (f-l)-^(y-l).  5.  (x^-f)-^(x-y), 

2.  (b'-125)-^(b-5).  6.  (a'-l)-^(a-l). 

3.  (a^  — 216)-f-(rt  — 6).  7.  (1  -  S;^^)  ^  (1  —  2cc). 

4.  (x^  —  S4:S)^(x  —  7).  8.  (x'  —  S2b')-i-(x  —  2b). 

9.  (8aV  — l)-f-(2aa:  — 1). 

10.  (l-27rr3«/^-T-(l-3xy). 

11.  (Ua%'  —  27a^-^(4:ab  —  3x), 

12.  (243a*-l)-J-(3a~l). 

13.  (32a*  — 243^*)-v-(2a-36). 

Case  II. 

(3)  ^-i-^  =  a^-a«^  +  «'^'-«^''  +  ^'. 
^  ^     a-f-b 

(4)  ^'^Q "^^  "to^^ ^  =  81  x^  -  54 x'lj  +  36 a^y -  24 a;/  + 16 1/. 

ox  ~j~  Zy 

From  these  results  it  may  be  assumed  that : 

103.  The  sum  of  two  equal  odd  powers  of  two  numbers  is 
divisible  by  the  sum  of  the  numbers. 

The  quotient  may  be  found  as  in  Case  I.,  but  the  signs 
are  alternately  plus  and  minus. 


56  ALGEBRA. 

Exercise  25. 

Write  by  inspection  the  results  in  the  following  exam- 
ples : 

1.  (x^^if)-^(x-^y).  5.  (8aV  +  l)-f-(2ax  +  l). 

2.  (^:^Jry')^{x^y).  6.  {x'^21  if)  ^  {x-\-^y). 

3.  (l  +  8a^)---(l  +  2a).  7.  (a*4-32&«)-f-(a-f  2^). 

4.  (27a'  +  ^'^)-T-(3a  +  ^).  8.  {^12xhf-{-z^-^{^xy-\-z). 

9.  (729a3  +  216^»^)-f-(9a  +  6^>). 

10.  (646^3+10005^) -^(4a  +  10^>). 

11.  (Q^a^b^-\-21x^)-^{4.ab^^x). 

12.  (x«  +  343)-^(a^  +  7). 

13.  {21xY-\-^z^)-^{^xy^2z). 

14.  (1024 a^  +  243 ¥)  -~(4.a^U). 

Case  III. 
(1)    ^Ef'^^  +  ^Z-  (2)    ^'  =  ^3  +  x^y  +  :r^  +  2^. 

(3)    ^=^-y-  (4)    ''^  =  x^-xhj  +  xf-,/. 

From  these  results  it  may  be  assumed  that  : 

104.  The  difference  of  two  equal  even  powers  of  two  num- 
bers is  divisible  by  the  difference  and  also  by  the  sum  of  the 
numbers. 

When  the  divisor  is  the  difference  of  the  numbers,  the 
quotient  is  found  as  in  Case  I. 

When  the  divisor  is  the  sum  of  the  numbers,  the  quo- 
tient is  found  as  in  Case  II. 


DIVISION.  67 


Exercise  26. 


Write  by  inspection  the  results  in  the  following  exam- 
ples : 

1.  {x'  —  y^)-^(x  —  y).  8.  (16a;*  — 1) -f- (2ic  +  l). 

2.  {x^  —  y^)^(x-{-y).  9.  (81aV— l)-^(3a.T  — 1). 

3.  {a^  —  x^-^(a  —  x).  10.  (81aV— l)-i- (3aic  +  l). 

4.  (r/  — a;«)-f- (a  +  a:).  11.  (64  «^  —  Z*«) -r- (2  a  —  J). 

5.  (a;*-8l2/*)-v-(a--37/).  12.  (64a«-^»«)^(2a  +  *). 

6.  (ic*-81z/^)-^(x  +  3y).  13.  (a;«— 7292/«)-j-(a:  — 3y). 

7.  (16a;*  — 1)^(20-  — 1).  14.  (a;«  — 729y«) -j- (x  +  Sy). 

15.  (81«*  — 16c*)-v-(3a  — 2  c). 

16.  (81«*  — 16c*)-v-(3a  +  2c). 

17.  (256  a*  — 10,000) -V- (4  «- 10). 

18.  (256«*  — 10,000) -r- (4  a +  10). 

19.  (625a-*  — l)-^(oa;  — 1). 

Case  IV. 
It  may  be  easily  verified  that : 

105.  The  sum  of  two  equal  even  powers  of  two  numhers 
is  not  divisible  by  either  the  sum  or  the  difference  of  the 
numbers. 

But  when  the  exponent  of  each  of  the  two  equal  powers 
is  composed  of  an  odd  and  an  even  factor,  the  sum  of 
the  given  powers  is  divisible  by  the  sum  of  the  powers 
expressed  by  the  even  factor. 

Thus,  x^  -\-  2/®  is  not  divisible  by  a;  +  ?/  or  by  ic  —  i^,  but  is 
divisible  by  x^  -f  y'^. 

The  quotient  may  be  found  as  in  Case  II. 


68  ALGEBRA. 


Exercise  27. 

Write  by  inspection  the  results  in  the  following  exam- 
ples : 

1.  (x'-\-y')-^{x'-\-f).  6.  (^12  +  1) -(^''  +  1). 

2.  (a«  +  l)-v-(a2  +  l).  7.  (64^«  +  7/«) -^  (4x2  +  2/2). 

3.  (aio  +  2/'°)^(^'  +  2/')-  8.  (64 +  ««) -^  (4  + a2^J. 

4.  (^»i"  + 1)  H-  {b''  + 1).  9.  (729 a^  +  Z*^)  -f-  (9 ^2+  ^*2^^. 

5.  {a^^-\-h'^)-^{a''-\-hy  10.  (729c'^  +  l)^  (96-2  +  1). 

Note.  The  introduction  of  negative  numbers  requires  an  exten- 
sion of  the  meaning  of  some  terms  common  to  arithmetic  and 
algebra.  But  every  such  extension  of  meaning  must  be  consistent 
with  the  sense  previously  attached  to  the  term  and  with  general  laws 
already  established. 

Addition  in  algebra  does  not  necessarily  imply  augmentation^  as  it 
does  in  arithmetic.  Thus,  7  +  (—  5)  =  2.  The  word  sum,  however, 
is  used  to  denote  the  result. 

Such  a  result  is  called  the  algebraic  sum,  when  it  is  necessary  to 
distinguish  it  from  the  arithmetical  sum,  which  would  be  obtained  by 
adding  the  absolute  values  of  the  numbers. 

The  general  definition  of  Addition  is,  the  operation  of  uniting 
two  or  more  numbers  in  a  single  expression  written  in  its  simplest 
form. 

The  general  definition  of  Subtraction  is,  the  operation  of  finding 
from  two  given  numbers,  called  minuend  and  subtrahend,  a  third 
number,  called  difference,  which  added  to  the  subtrahend  will  give  the 
minuend. 

The  general  definition  of  Multiplication  is,  the  operation  of  finding 
from  two  given  numbers,  called  multiplicand  and  multiplier,  a  third 
number,  called  product,  which  may  be  formed  from  the  multiplicand 
as  the  multiplier  is  formed  from  unity. 

The  general  definition  of  Division  is,  the  operation  of  finding  the 
other  factor  when  the  product  of  two  factors  and  one  factor  are 
given. 


CHAPTER   y. 

Simple  Equations. 

106.  An  equation  is  a  statement  that  two  expressions 
are  equal.     Thus,  4  a;  — 12  =  8. 

107.  Every  equation  consists  of  two  parts,  called  the 
first  and  second  sides,  or  members,  of  the  equation. 

108.  An  identical  equation  is  one  in  which  the  two  sides 
are  equal,  whatever  numbers  the  letters  stand  for.     Thus, 

(x-\-h){x  —  h)=::X^  —  h\ 

109.  An  equation  of  condition  is  one  which  is  true  only 
when  the  letters  stand  for  2>ci^'ticuLar  values.  Thus,  ic  +  5 
=  8  is  true  only  when  a;  =  3. 

110.  A  letter  to  which  a  particular  value  must  be  given 
in  order  that  the  statement  contained  in  an  equation  may 
be  true  is  called  an  unknoivn  niunher. 

111.  The  value  of  the  unknown  number  is  the  number 
which  substituted  for  it  will  satisfy  the  equation,  and  is 
called  a  root  of  the  equation. 

112.  To  solve  an  equation  is  to  find  the  unknown  num- 
ber. 

113.  A  simple  equation  is  one  which  contains  only  the 
first  power  of  the  symbol  for  the  unknown  number,  and  is 
also  called  an  equation  of  the  first  degree. 


60  ALGEBRA. 

114.  If  equal  changes  he  made  in  both  sides  of  an  equa- 
tion, the  results  will  he  equal.  §  43. 

(1)  To  find  the  value  oi  x  in  x -\- h  =^  a. 

x  +  6  =  a; 
Subtract  6  from  each  side,       x  +  h  —  h— a  —  h; 
Cancel  +  6  —  5,  x=  a  —  h. 

(2)  To  find  the  value  of  a;  in  cc  —  h  =  a. 

x  —  b=  a; 
Subtract  —  h  from  each  side,  x  —  h+b=a-{-b; 
Cancel  — 6  +  6,  x  =  a  +  b. 

The  result  in  each  case  is  the  same  as  if  b  were  trans- 
posed to  the  other  side  of  the  equation  with  its  sign 
changed.     Therefore, 

115.  Any  term  may  he  transposed  from  one  side  of  an 
equation  to  the  other  provided  its  sign  he  changed. 

For,  in  this  transposition,  the  same  number  is  subtracted 
from  each  side  of  the  equation. 

116.  The  signs  of  all  the  terms  on  each  side  of  an 
equation  may  be  changed ;  for  this  is  in  effect  transposing 
every  term. 

117.  When  the  symbols  for  the  known  and  unknown 
numbers  of  an  equation  are  connected  by  the  sign  +  or  — , 
they  may  be  separated  by  transposing  the  symbols  for 
the  known  numbers  to  one  side  and  the  symbols  for  the 
unknown  to  the  other. 

118.  Hence,  to  solve  an  equation  with  one  unknown 
number, 

Transpose  all  the  terms  involving  the  unknown  numher  to 
the  left  side,  and  all  the  other  terms  to  the  right  side;  combine 


SIMPLE    EQUATIONS.  61 

the  like  terms,  and  divide  both  sides  by  the  coefficient  of  the 
sywhol  for  the  unknown  number. 

119.    To  verify  the  result,  substitute  the  value  of  the 
symbol  for  the  unknown  number  in  the  original  equation. 


Exercise  28. 
Find  the  value  of  x  in 

1.  5a;  — 1  =  19.  8.  16a; -11  =  7a; +  70. 

2.  3a;  +  6  =  12.        _  9.  24a;  — 49  =  19a;  — 14. 

3.  24a;  =  7a; -f- 34.  10.  3a; +  23  =  78  — 2a;. 

4.  8a;  — 29  =  26  — 3a;.  11.  26  — 8a;  =  80  — 14a;. 


..    12— 5a;  r=  19  — 12a;.  12.    13  — 3a;  =  5a;  — 3. 

;.    3a;  +  6-\2a;  =  7a;.  13.    3a;  — 22  =  7a;  +  6. 


^    7.    5a; +  50  =  4a; +  56.  14.    8  +  4a- =  12a;  — 16. 

16.  5a;  — (3a;  — 7)=  4x  — (6a;  — 35). 

16.  6a;  — 2(9— 4a;) +  3 (5a;  — 7)  =  10a;  — (4  + 16a; +  35). 

17.  9a;  — 3(5a;  — 6)  +  30  =  0. 

18.  a;  — 7(4a;-ll)  =  14(a;  — 5)-19(8  — a;)  — 61. 

19.  (a;  +  7)(a;  -  3)  =  (a; -5)(a;- 15). 

20.  (a;  — 8)  (a; +  12)  =  (a; +  1)  (a;  — 6). 

21.  (a;-2)(7-a;)  +  (a;-5)(.T  +  3)-2(a;-l)+12  =  0. 

22.  (2a;-7)(a;  +  5)  =  (9-2a-)(4-^)  +  229. 

23.  14  — a;  — 5(a;— 3)(a;  +  2)  +  (5  — a;)(4  — 5a;)  =  45a;  — 76. 

24.  (x  +  5y-(4.-xy  =  21x. 

25.  5(a;  — 2)'^  +  7(a;-3)2=(3a;-7)(4a;-19)  +  42. 
V 


ALGEBRA. 


Exercise  29. 

problems. 

1.  Find  a  number  such  that  when  12  is  added  to  its  double 

the  sum  shall  be  28. 

Let  X  =  the  number. 

Then  2  x  =  its  double, 

and        2  cc  +  12  =  double  the  number  increased  by  12. 
But  28  =  double  the  number  increased  by  12. 

.-.  2ic+ 12  =  28. 

2ic=28-12. 
2x=  16, 
x=8. 

2.  A  farmer  had  two  flocks  of  sheep,  each  containing  the 

same  number.  He  sold  21  sheep  from  one  flock  and 
70  from  the  other,  and  then  found  that  he  had  left  in 
one  flock  twice  as  many  as  in  the  other.  How  many 
had  he  in  each? 

Let  X  =  the  number  of  sheep  in  each  flock. 

Then     x  —  21  =  the  number  of  sheep  left  in  one  flock, 
and  x~  70  =  the  number  of  sheep  left  in  the  other. 

.-.  x-21  =  2(x-70), 
X  —  21  =  2  X  -  140. 
x-2x=  -140  +  21, 
-x=  -119, 
X  =  119. 

3.  A  and  B  had  equal  sums  of  money ;  B  gave  A  $5,  and 

then  3  times  A's  money  was  equal  to  11  times  B's 
money.     What  had  each  at  first  ? 

Let  X  =  the  number  of  dollars  each  had. 

Then  x  +  5  =  the  number  of  dollars  A  had  after  receiving 
$5  from  B, 
and         X  —  6  =  the  number  of  dollars  B  had  after  giving  A  $5. 


SIMPLE    EQUATIONS.  63 

.-.  3(x4-5)  =  ll(x-5); 
3x+15=llx-65; 
3x  — llx=  —  55  — 15; 

-8x=-70; 
x=8f. 

Therefore,  each  had  $8.75. 

4.   Find  a  number  whose  treble  exceeds  50  by  as  much  as 
its  double  falls  short  of  40. 

Let  X  =  the  number. 

Then  3  x  =  its  treble, 

and        3  X  —  50  =  the  excess  of  its  treble  over  50  ; 
also,       40  —  2  X  =  the  number  its  double  lacks  of  40. 
.-.  3  X  —  50  =  40  —  2  X  ; 
3x  +  2x  =  40  +  50; 
6x=90; 
x=18. 

6.    What  two  numbers  are  those  whose  difference  is  14, 
and  whose  sum  is  48? 

Let  X  =  the  larger  number. 

Then         48  —  x  =  the  smaller  number, 
and     X  —  (48  —  x)  =  the  difference  of  the  numbers. 
But  14  =  the  difference  of  the  numbers. 

.•:x  -  (48  -  X)  =  14 
X  -  48  +  X  =  14 
2x  =  62 
x=31. 

Therefore,  the  two  numbers  are  31  and  17. 

6.  To  the  double  of  a  certain  number  I  add  14,  and  obtain 

as  a  result  154.     What  is  the  number  ? 

7.  To  four  times  a  certain  number  I  add  16,  and  obtain  as 

a  result  188.     What  is  the  number  ?  ^^^^ 

/  8.  By  adding  46  to  a  certain  number,  I  obtain  as  a  result 
a  number  three  times  as  large  as  the  original  number. 
Find  the  original  number. 


64  ALGEBRA. 

9.  One  number  is  three  times  as  large  as  another.  If  I 
take  the  smaller  from  16  and  the  greater  from  30, 
the  remainders  are  equal.     What  are  the  numbers? 

10.  Divide  the  number  92  into  four  parts,  such  that  the 

first  exceeds  the  second  by  10,  the  third  b}^  18,  and 
the  fourth  by  24. 

11.  The  sum  of  two  numbers  is  20 ;  and  if  three  times  the 

smaller  number  is  added  to  five  times  the  greater, 
the  sum  is  84.     What  are  the  numbers  ? 

12.  The  joint  ages  of  a  father  and  son  are  80  years.    If  the 

age  of  the  son  were  doubled,  he  would  be  10  years 
older  than  his  father.     What  is  the  age  of  each  ? 

13.  A  man  has  6  sons,  each  4  years  older  than  the  next 

younger.  The  eldest  is  three  times  as  old  as  the 
youngest.     What  is  the  age  of  each  ? 

14.  Add  $24  to  a  certain  sum  and  the  amount  will  be  as 

much  above  $80  as  the  sum  is  below  $80.  What 
is  the  sum  ? 

15.  Thirty  yards  of  cloth  and  40  yards  of  silk  together 

cost  $330;  and  the  silk  cost  twice  as  much  a  yard 
as  the  cloth.     How  much  did  each  cost  a  yard  ? 

16.  Find  the  number  whose  double  increased  by  24  exceeds 

80  by  as  much  as  the  number  itself  is  less  than  100. 

17.  The  sum  of  $500  is  divided  among  A,  B,  C,  and  D.     A 

and  B  have  together  $280,  A  and  C  $260,  and  A 
and  D  $220.     How  much  does  each  receive  ? 

18.  In  a  company  of  266  persons  composed  of  men,  women, 

and  children,  there  are  twice  as  many  men  as  women, 
and  twice  as  many  women  as  children.  How  many 
are  there  of  each? 


y 


SIMPLE    EQUATIONS.  65 

19.  rind  two  numbers  differing  by  8,  such  that  four  times 

the  less  may  exceed  twice  the  greater  by  10. 

20.  A  is  58  years  older  than  B,  and  A's  age  is  as  much 

above  60  as  B's  age  is  below  50.  Find  the  age  of 
each. 

21.  A  man  leaves  his  property,  amounting  to  $7500,  to  be 

divided  among  his  wife,  his  two  sons,  and  three 
daughters,  as  follows  :  a  son  is  to  have  twice  as 
much  as  a  daughter,  and  the  wife  $500  more  than 
all  the  children  together.  How  much  is  the  share 
of  each  ? 

22.  A  vessel  containing  some  water  was  filled  by  pouring 

in  42  gallons,  and  there  was  then  in  the  vessel  seven 
times  as  much  as  at  first.  How  much  did  the  vessel 
hold? 

23.  A  has  $72  and  B  has  $52.     B  gives  A  a  certain  sum ; 

then  A  has  three  times  as  much  as  B.  How  much 
did  A  receive  from  B  ? 

24.  Divide  90  into  two  such  parts  that  four  times  one  part 

may  be  equal  to  five  times  the  other. 

25.  Divide  60  into  two  such  parts  that  one  part  exceeds 

the  otlier  by  24. 

26.  Divide  84  into  two  such  parts  that  one  part  may  be 

less  than  the  other  by  36. 

Note  I.  When  we  have  to  compare  the  ages  of  two  persons  at  a 
given  time,  and  also  a  number  of  years  after  or  before  the  given  time, 
we  must  remember  that  both  persons  will  be  so  many  years  older  or 
younger. 

Thus,  if  X  represent  A's  age,  and  2x  B's  age,  at  the  present  time, 
A's  age  five  years  ago  will  be  represented  by  x  —  5  ;  and  B's  by 
2  X  —  5.  A's  age  five  years  hence  will  be  represented  by  x  +  5  ;  and 
B's  age  by  2  X  +  5. 


66  ALGEBRA. 

27.  A  is  twice  as  old  as  B,  and  22  years  ago  he  was  three 

times  as  old  as  B.     What  is  A's  age? 

28.  A  father  is  30  and  his  son  6  years  old.     In  how  many 

years  will  the  father  be  just  twice  as  old  as  the  son  ? 

29.  A  is  twice  as  old  as  B,  and  20  years  ago  he  was  three 

times  as  old.     What  is  B's  age  ? 

30.  A  is  three  times  as  old  as  B,  and  19  years  hence  he 

will  be  only  twice  as  old  as  B.  What  is  the  age  of 
each? 

31.  A  man  has  three  nephews  ;  his  age  is  50,  and  the  joint 

ages  of  the  nephews  is  42.  How  long  will  it  be 
before  the  joint  ages  of  the  nephews  will  be  equal 
to  that  of  the  uncle  ? 

Note  II.  In  problems  involving  quantities  of  the  same  kind 
expressed  in  different  units,  we  must  be  careful  to  reduce  all  the 
quantities  to  the  same  unit. 

Thus,  if  X  denotes  a  number  of  inches,  all  the  quantities  of  the  same 
kind  involved  in  the  problem  must  be  reduced  to  inches. 

32.  A  sum  of  money  consists  of  dollars  and  twenty-five- 

cent  pieces,  and  amounts  to  $20.  The  number  of 
coins  is  50.     How  many  are  there  of  each  sort? 

33.  A  person  bought  30  pounds  of  sugar  of  two  different 

kinds,  and  paid  for  the  whole  $2.94.  The  better 
kind  cost  10  cents  a  pound  and  the  poorer  kind  7 
cents  a  pound.  How  many  pounds  were  there  of 
each  kind? 

34.  A  workman  was  hired  for  40  days,  at  $1  for  every  day 

he  worked,  but  with  the  condition  that  for  every 
day  he  did  not  work  he  was  to  pay  45  cents  for  his 
board.  At  the  end  of  the  time  he  received  $22.60. 
How  many  days  did  he  work? 


SIMPLE    EQUATIONS.  67 

35.  A  wine  merchant  has  two  kinds  of  wine;  one  worth  50 

cents  a  quart,  and  the  other  75  cents  a  quart.  From 
these  he  wishes  to  make  a  mixture  of  100  gallons, 
worth  $2.40  a  gallon.  How  many  gallons  must  he 
take  of  each  kind? 

36.  A  gentleman  gave  some  children  10  cents  each,  and 

had  a  dollar  left.  He  found  that  he  would  have 
required  one  dollar  more  to  enable  him  to  give  them 
15  cents  each.     How  many  children  were  there  ? 

37.  Two  casks  contain  equal  quantities  of  vinegar ;  from 

the  first  cask  34  quarts  are  drawn,  from  the  second, 
20  gallons ;  the  quantity  remaining  in  one  vessel  is 
now  twice  that  in  the  other.  How  much  did  each 
cask  contain  at  first? 

38.  A  gentleman  hired  a  man  for  12  months,  at  the  wages 

of  $90  and  a  suit  of  clothes.  At  the  end  of  7 
months  the  man  quits  his  service  and  receives 
$33.75  and  the  suit  of  clothes.  What  was  the  price 
of  the  suit  of  clothes  ? 

39.  A  man  has  three  times  as  many  quarters  as  half- 

dollars,  four  times  as  many  dimes  as  quarters,  and 
twice  as  many  half-dimes  as  dimes.  The  whole  sum 
is  $7.30.     How  many  coins  has  he  in  all  ? 

40.  A  person  paid  a  bill  of  $15.25  with  quarters  and  half- 

dollars,  and  gave  51  pieces  of  money  all  together. 
How  many  of  each  kind  were  there  ? 

41.  A  bill  of  100  pounds  was  paid  with  guineas  (21  shil- 

lings) and  half-crowns  (2^  shillings),  and  48  more 
half-crowns  than  guineas  were  used.  How  many  of 
each  were  paid  ? 


CHAPTER   VI. 
Factors. 

120.  An  expression  is  rational  if  none  of  its  terms  con- 
tain square  or  other  roots.  It  is  often  important  to 
determine  the  rational  and  integral  factors  of  a  given 
expression. 

121.  Case  I.  The  simplest  case  is  that  in  which  all  the 
terms  of  an  expression  have  one  common  factor.     Thus, 

(1)  x^-\-xy  =  x{x-]-y). 

(2)  6a3  +  4a2  +  8a  =  2a(3a2  +  2a  +  4). 

(3)  ld>a^b  —  21a^b^-\-ZQah  =  'dah{2a^  —  3ah-\-4:). 

Exercise  30. 
Eesolve  into  factors : 

1.  5^2  — 15a.  4.    4a;V  — 12xy  +  8V- 

2.  6a3  +  18a2  — 12a.  5.    y^  —  af^hf-^-cy. 

3.  49x2  — 21  a^  + 14.  6.    Qa''b^  —  21a^V'-{'21a%\ 

7.  54icy  +  108£cy  — 243rry. 

8.  45  a; yo  —  90  xhf  —  360  ^y . 

9.  70  ay -140  ay +  210  a/. 
10.  32a«//  +  96a«^»«-128a8^». 


FACTORS.  69 

122.  Case  II.  Frequently  the  terms  of  an  expression 
can  be  so  arranged  as  to  show  a  common  factor.     Thus, 

(1)  x^-{-ax-i-bx-\-ab  =  (x^  +  ax)  +  (bx  +  ab), 

=  x(x-\-a)-\-b(x-\-a), 
=  (x-\-b)(x-}-a). 

(2)  ac  —  ad  —  bc-\-bd=  (ac  —  ad)  —  (be  —  bd), 

^a(c  —  d)  —  b(c  —  d)y 
=  {a  —  b){c  —  i). 

Exercise  31. 
Resolve  into  factors  : 

1.  x^  —  ax  —  bx-\-ab.  6.  abx  —  aby-{-pqx — pqy. 

2.  ab-j-ay  —  by  —  if.  7.  cdx^-\-adxy  —  bcxy  —  abif. 

3.  be -\- bx  —  ex  —  x^.  8.  abey  —  bHy  —  acdx -\- bd^x. 

4.  mx -\- vin -\- ax -\- an.  9.  ax  —  ay  —  bx-\-by. 

5.  edx^  —  cxy-\-dxy  —  if.  10.  edz^  —  eyz -\- dyz  —  \f . 

123.  The  square  root  of  a  number  is  one  of  the  two  equal 
factors  of  that  number.  Thus,  the  square  root  of  25  is  5  ; 
for,  25  =  5X5. 

The  square  root  of  a*  is  a^ ;  for,  a^  =  a^  X  a\ 

The  square  root  of  a^b^c^  is  abc ;  for,  a^^V  =  abe  X  abe. 

In  general,  the  square  root  of  a  power  of  an  expression 
is  found  by  taking  the  square  root  of  the  numerical  co- 
efficient and  writing  each  letter  with  an  exponent  equal  to 
one-half  the  exponent  of  the  letter  in  the  given  power. 

The  square  root  of  a  product  may  be  found  by  taking 
the  square  root  of  each  factor,  and  finding  the  product  of 
the  roots. 


70  ALGEBRA. 


The  square  root  of  a  positive  number  may  be  either 
positive  or  negative;  for, 


a?^a  X  a, 


or. 


but  throughout  this  chapter  only  the  positive  value  of  the 
square  root  will  be  taken. 

124.  Case  III.  From  §  73  it  is  seen  that  a  trinomial  is 
often  the  product  of  two  binomials.  Conversely,  a  trino- 
mial may,  in  certain  cases,  be  resolved  into  two  binomial 
factors.     Thus, 

To  find  the  factors  of 

ic2  4-7£c+12. 

The  first  term  of  each  binomial  factor  will  obviously  be  x. 

The  second  terms  of  the  two  binomial  factors  must  be 
two  numbers 

whose  product  is  12, 
and  whose  sum  is         7. 

The  only  two  numbers  whose  product  is  12  and  whose 
sum  is  7  are  4  and  3. 

.'.x^-\-7x-\-12={x-{-4:)(x-{-S). 

Again,  to  find  the  factors  oi  x^-}-5xi/-\-  6y^. 

The  first  term  of  each  binomial  factor  will  obviously  be  x. 

The  second  terms  of  the  two  binomial  factors  must  be 
two  numbers 

whose  product  is  6 1/^, 
and  whose  sum  is        5j/. 

The  only  two  numbers  whose  product  is  6y^  and  whose 
sum  is  5y  are  3y  and  2y. 

.•.x'-\-5xy  +  6f=(x-JrSy)(x-}-2yy 


FACTORS.  71 

Exercise  32. 
Find  the  factors  of  : 

1.  x'  +  llx  +  24:.  11.  x^-i-lSax-{-S6a\ 

2.  ir;+llic  +  30.  12.  f-{-19pi/  +  4:Sp\ 

3.  2/2  _^  17?/ +  60.  13.  z^ -\- 29  qz-\- 100  q\ 

4.  ^2_^13^  +  12.  14.  a^  +  5a2-f6. 

5.  a^2^21:r  +  110.  15.  ^"  +  4^  +  3. 

6.  y2  +  35y  +  300.  16.  a2^2_j_;^g^j_|.32, 

7.  ^»2_|.23^  +  102.  17.  a:y  +  7a;y  +  12. 

8.  a:2  +  3a-  +  2.  18.  ^»"  +  10;5^  + 16. 

9.  ic2  +  7cc  +  6.  19.  a2_^9a^>  +  20^2^ 
10.    a^-i-9ab-\-Sb\  20.  ic«  +  9a;3  +  20. 

21.  a^x^-\-14:abx-\-S3b\         24.    ^'V+ 18 aZ»c  + 65 al 

22.  aV  + 7  aca^+ 10 a;l  25.    ? V  +  23  rs;^  +  90 ;s;l 

23.  a;y^2_^19a;?/^  +  48.         26.    mV  +  20  ;/iV/;j/ +  51  j^V- 

126.   Case  IV.    To  find  the  factors  of  (/ 

a;2  — 9a;  +  20. 


The  second  terms  of  the  two  binomial  factors  must  be 
two  numbers 

whose  product  is  20, 
and  whose  sum  is       —  9. 

The  only  two  numbers  whose  product  is  20  and  whose 
sum  is  —  9  are  —  5  and  —  4. 

,'.x^  —  9x-\-20  =  (x  —  5)(x  —  4:). 


72  ALGEBRA. 

Exercise  33. 


Eesolve  into  factors 


V 


1.  ic2  — 7x  +  10.  13.  w'b^c^  — IS abc  + 22. 

2.  x^  —  29x  +  190.  14.  0^2  —  150^  +  50. 

3.  a^  —  23a  +  lS2.  15.  ic2_20ic  +  100. 

4.  ^'2_30^-f  200.  16.  aV  — 21ao:  +  54. 

5.  ^2_43^_|_46o.  17.  aV-16a^»o^  +  39^>^ 

6.  o;2  — 7o:  +  6.  18.  a^c^  —  24:acz-\-14:Sz\ 

7.  o;^  — 4aV  +  3a^  19.  rc^  — 20o;  +  91. 

8.  o;2_8a._|_i2.  20.  ic^  —  23 a:  + 120. 

9.  z^  — 57^  +  56.  21.  ^2_53^_^3(30. 

10.  ?/''  — 7y^+12.  22.  .T^— (a  +  c)o;  +  ac. 

11.  a;y-27ic?/  +  26.  23.  ifz' -  28  ahjz -\- 187 a'b\ 

la' 


12.    a*b^  — 11  a%^  + 30.  24.    c2(^2_30a^C£^_|_221a2^2_ 


126.    Case  V.    To  find  the  factors  of 

a^  +  2ic  — 3. 

The  second  terms  of  the  two  binomial  factors  must  be 
two  numbers 

whose  product  is  —  3, 
and  whose  sum  is        +  2. 

The  only  two  numbers  whose  product  is  —  3  and  whose 
sum  is  +  2  are  +  3  and  —  1. 

,'.x'-\-2x  —  3  =  (x  +  3)(x  —  l). 


FACTORS.  73 

Exercise  34. 

Resolve  into  factors  :  ,  ^ 

1.  x^-]-6x  —  7.  8.  a'  +  25a  —  150, 

2.  x'-{-5x-S^.  9.  b'-{-Sb'-4.. 

3.  if+7ij  —  60.  10.  iV  +  3^c  — 154. 

4.  if -\- 12  2j- 4:5.  11.  c^°  +  15c^-100. 

5.  z'-\-llz  —  12.  12.  c2  +  17c  — 390. 

6.  z'-\-lSz-UO.  13.  a2+a-132. 

7.  a^-]- 13  a  — 300.  \J   14.  icy^=^  +  9xy;2-22. 

127.   Case  VI.    To  find  the  factors  of 

x^  —  5x  —  66. 

The  second  terms  of  the  two  binomial  factors  must  be 
two  numbers 

whose  product  is  —  66, 
and  whose  sum  is       —   5. 

The  only  two  numbers  whose  product  is  —  66  and  whose 
sum  is  —  5  are  — 11  and  +  6. 

.\x^  —  5x  —  66  =  (x  —  ll)(x-i-6). 

Exercise  35. 

Resolve  into  factors  : 

'^.    ic2  — 3ic  — 28.  6.  a^  — 15a  — 100. 

2.  2/2-7?/-18.  7.  c^«-9c^-10. 

3.  ^2  — 9a;  — 36.  8.  aj2_8a:  — 20. 

4.  z'  —  llz  —  60.  9.  f  —  5ay  —  50a^. 

5.  z^  —  13z  —  14:.  10.  a^^  —  3ab  —  4:. 


74  ALGEBRA. 

11.  aV  — 3acc  — 54.  14.   y^^"  -  5  2/V  —  84. 

12.  c2^  — 24c^  — 180.  15.    a2^>2— iGa^*  — 36. 

13.  aV  — a^c  — 2.  16.    x^— (a  — ^»)a;  — aJ. 

We  now  proceed,  to  the  consideration  of  trinomials  which 
are  perfect  squares.  These  are  only  particular  forms  of 
Cases  III.  and  IV.,  but  from  their  importance  demand 
special  attention. 

128.   Case  VII.     To  find  the  factors  of 

a;2  +  18a;  +  81. 

The  second  terms  of  the  two  binomial  factors  must  be 
two  numbers 

whose  product  is  81, 
and  whose  sum  is        18. 

The  only  two  numbers  whose  product  is  81  and  whose 
sum  is  18  are  9  and  9. 

.'.x'-\-l^x  +  Sl  =  (x-\-9)(x-\-^)  =  {x-\-^y. 


Exercise  36. 
Resolve  into  factors : 

1.  ir2_|_i2x  +  36.  8.  y^-\-l(Sfz^-\-Q4.z\ 

2.  a;2  +  28x  +  196.  9.  i/-{-2i:tf-\-14.L 

3.  cc^ +  34^^  +  289.  10.  x'z'-\-lQ2xz-\-mQ>l. 

4.  z'^^z-^l.  11.  4.a^-\-12ab^^^h\ 

5.  /  + 200  2/ +  10,000.  12.  9xh/-\-3^xifz-]-25z\ 

6.  ;?*  +  14^2  +  49.  13.  <dx^-\-12xy-{-4.f. 

7.  x^-\'S6xy  +  324:f.  14.  4aV  +  20aVy  +  25xy. 


FACTORS. 


75 


129.   Case  VIII.     To  find  the  factors  of 

x2  — 18a;  +  81. 
The  second  terms  of  the  two  binomials  must  be  two 


numbers 
and 


whose  product  is  81, 
whose  sum  is    — 18. 


The  only  two  numbers  whose  product  is  81  and  whose 
sum  is  —18  are  —9  and  —9. 


.2 


18x  +  81  =  (x  — 9)(x  — 9)  =  (x  — 9)2. 


Exercise  37. 


Resolve  into  factors 

1.  a'  —  ^a-\-l&. 

2.  a2_30rt-j_225. 

3.  ir2-38x  +  361. 

4.  £c2_4o^^400. 

5.  y2  _  100 y  + 2500. 

6.  y'  — 20  if +  100. 

7.  7/2 -50  2/^4- 625^2. 

8.  a;^-32icy  +  2562/^ 

9.  ^«  — 34^»  +  289. 


10.  4a;y  — 20iry^  +  25yV. 

11.  l^xh/  —  ^xifz''-\-fz\ 

12.  9a2^,V-6aZ.V(i  +  6V<^. 

13.  16a:«— 8xy  +  a;y. 

14.  aV  — 2a«6xy  +  ^»y. 

15.  36xy  — 60a;7/«  +  25y. 

16.  l-Q,ab^-\-9a%\ 

17.  9mV  — 24wm  +  16. 

18.  Wx-  —  12hx'y\-^xY, 

19.  49^2  — 112aZ>  + 64^1 

20.  64  x^  — 160  xUf^  +  100  x^z\ 

21.  49a2^V  — 28«^'Cic  +  4ic2. 

22.  121cc^  — 286a:V  +  169y2 

23.  2^'da^fz'  —  102xfzH-\-^fzW. 

24.  361icy«2_jg^j^^y^^4^2^2^ 


76  ALGEBRA. 

130.  Case  IX.  An  expression  in  the  form  of  two 
squares,  with  the  negative  sign  between  them,  is  the  prod- 
uct of  two  factors  which  may  be  determined  as  follows  : 

Take  the  square  root  of  the  first  term,  and  the  square 
root  of  the  second  term. 

The  sum  of  these  roots  will  form  the  first  factor  ; 

The  difference  of  these  roots  will  form  the  second  factor. 
Thus: 

(1)  a'  —  h^={a-{-h){a  —  h). 

(2)  a'-(b-cy=.\a^{h-c)\\a-{b-c)\, 

(3)  {a-bf-{c-<^'^=\{a-h)-^(c-d)\\{a-b)-(c-d)l 

■    =\a  —  h-\-c  —  d\\a  —  h  —  c-\-d\. 

131.  The  terms  of  an  expression  may  often  be  arranged 
so  as  to  form  two  squares  with  the  negative  sign  between 
them,  and  the  expression  can  then  be  resolved  into  factors. 
Thus: 

a'^h''  —  c'  —  d''-\-2ab-\-2cd, 
=  a}^2ab-^b^  —  c^^2cd  —  d\ 
=={a'-^2ab-\-b')-{c'  —  2cd^d^, 
=  i^aJ^bf-{c-df, 

=  \ia^b)-\-{c-d)\\(a-^b)-{c-d)\, 
=  \a^b^c  —  d\\a-\-b  —  c-\-d\. 

132.  An  expression  may  often  be  resolved  into  three  or 
four  factors.     Thus : 

(1)  x''-if'={x'-\-f)(x'-f) 

=  (x'-\-f){x'^-y'){x'^-f)(x^y)(x-y). 


FACTORS.  77 

(2)  4:(ab-i-cdy—(a'  +  b'  —  c^  —  dy, 
==\2(ab-^cd)-\-(a'-\-b'  —  c'-d^\ 

\2(ab-i-cd)-(a'-}-b'-c'-d^\, 
=  \2ab-i-2cd-{-a'-{-b'  —  c'~d'\ 

\2ab  +  2cd-a'--b'-^c^-{-d^, 
=  \(a'  +  2ab  +  b^  —  (c'-2cd-{-d')l 

\(c'-\-2cd-i-d^-~(a'-2ab-{-b'^l, 

=  \(a-{-by-(c-dyi\(c  +  dy-(a-byi, 

=  \a  +  b-{-(c  —  d)\\a-\-b  —  (c  —  d)\ 

lc  +  d-}-(a-b)\\c  +  d-'(a-b)l 
=  \a  +  b  +  c  —  dl\a-\'b—c-{-d\ 

Ic-^-d+a  —  bllc  +  d  —  a  +  bl 


Exercise  38. 

Resolve  into  factors  : 

1.  a^  —  b-.  14,  (a-{-by—(c-\-dy. 

2.  a^  —  16.  15.  (x-\-i/y—(x  —  yy, 

3.  4^2  — 25.  16.  2ab  —  a^  —  b'i-l. 

4.  a*~b\  17.  x^  —  2yz  —  f  —  z^. 

5.  a^  —  \.  18.  x''—2xy-^if—z^. 

6.  a^  —  b\  19.  ie^Vlbc  —  W  —  <^(^. 

7.  a»  — 1.  20.  a2_2a^-[-y2_^2_2a;;^  — ;^2. 

8.  36x2--492/2.  21.  2 xy  —  x^  —  ?/-' +  ^l 

9.  100 xy  — 121^262.  22.  a:2+7/^— ^2— (^2_2a:y_2c?;s. 

10.  1  —  49x2.  23.  x^—if^z''—a}—2xz-^2ay. 

11.  a*  —  25b^.  24.  2a^  +  a'  +  ^'  — c^. 

12.  (a--^»)2  — cl  25.  2x2/  — X-  — y2  +  a2+Z»2  — 2a^. 

13.  x^-'(a  —  by.  26.  (ax  +  ^'y)^  — 1. 


ji  ALGEBRA. 

27.  1— x2  — ?/2_|_2a.y.  31.  (a;  + l)^— (y  — 1)2. 

28.  (5a  —  2y~(a  —  4.y.  32.  d^  —  x^ -{- Axy  -  4:f, 

29.  a'  —  2ab-\-b''  —  x'.  33.  a^  — j2_2^,c  — c^. 

30.  (x-\-iy-(i/  +  iy.  34.  4x^-9ic2  +  6x-l. 

133.  CaseX. 

/^3 ,^ 

Since  — -~-  ^=x^-\-xy-\-iy, 

x      y  , 

x^  —  1/* 

and  —  =  x*-\-  xhj  +  ic  ^  +  ^if  +  2/S 

and  so  on,  it  follows  that  the  difference  between  two  equal 
odd  powers  of  two  numbers  is  divisible  by  the  difference 
between  the  numbers. 

Exercise  39. 
Resolve  into  factors : 

1.  a^  —  b\  6.  8a;«  — 27?/^. 

2.  cc^  — 8.  7.  64^  — 1000 ^^ 

3.  ic^  — 343.  8.  729x^  —  5122/^. 

4.  7/^-125.  9.  27  a^- 1728. 

5.  if  —  2U.  10.  1000  a^- 1331  ^>«. 

134.  Case  XL 

01?  — l—  ^^ 
Since  — ; —  =  x^  —  ax'\-  a?, 

x-\-a 

X?  "4~  It* 
and  — -r-^  ^  a:;'*  —  a:;'V  +  icy  —  ic?/^  +  y*, 

and  so  on,  it  follows  that  the  sum  of  two  equal  odd  powers 
of  two  numbers  is  divisible  by  the  sum  of  the  numbers. 


FACTORS.  79 

Exercise  40. 
Resolve  into  factors  : 

1.  x^-^if.  6.  216tt»  +  512c^ 

2.  a;«  +  8.               ^  7.  729  a:«+ 17282/*. 

3.  x«  +  216.  ^  8.  x'-^if. 

4.  ?/3  +  64;^l  /:  9.  a;'  +  /. 

5.  646»+125c«.  /^lO.  32^»«  +  243c». 

135.  Case  XII.  The  sum  of  any  two  ^powers  of  two 
numbers,  whose  exponents  contain  the  same  odd  factor,  is 
divisible  by  the  sum  of  the  powers  obtained  by  dividing 
the  exponents  of  the  given  powers  by  this  odd  factor. 

Thus, 

In  like  manner,  rr^°  +  32y',  which  is  equal  to  x^^-\-  (2yyy 
is  divisible  by  a;*  -j-  2  y ;  but  x*  -\-  y*y  whose  exponents  do  not 
contain  an  odd  factor,  and  x^-\-y^^,  whose  exponents  do 
not  contain  the  same  odd  factor,  cannot  be  resolved  into 
rational  factors. 

Exercise  41. 
Resolve  into  factors  : 

1.  a^^b\  3.    ic"  +  y".  5.    a;«+l.  7.    64a«  +  a;«. 

2.  a^«  +  ^»^".       4.    d«  +  64c«.         6.    a^^-\-l.         8.    729 +  c«. 


80  ALGEBRA. 

136.  Case  XIII.  For  a  trinomial  to  be  a  perfect  square, 
the  middle  term  must  he  twice  the  product  of  the  square  roots 
of  the  first  and  last  terms. 

The  expression  x^ -\- x^y^ -\- y*^  will  become  a  perfect 
square  if  a^y  be  added  to  the  middle  term.  And  if  the 
subtraction  of  a^y  from  the  expression  thus  obtained  be 
indicated,  the  result  will  be  the  difference  of  two  squares. 
Thus: 

=  {x'-^yy-xhf, 
=  {x^-\-y^-\-xy){x^-\rf  —  xy), 
or,  (x^  -\-xy-["if)  (x^  —  xy-\-  y^. 


Exercise  42. 
Resolve  into  factors  : 

1.  a''-\-a''h''-\-h\  8.  49m*  +  110wV  +  8l7i*. 

2.  9x*  +  3a;y  +  42/*.  9.  9a*  +  21aV  +  25c*. 

3.  16x^  — 17xy  +  2/'-  10.  ^9a'  —  lba%^-^121b\ 
4..  Sla^-\-2^a^b^^lU\  11.  64x^+128ccy +  8I2/*. 

5.  ^la^  —  2%a''b^-\-lU\  12.  *4aj^- 37xy +  93/*. 

6.  9x^  +  38ccy  +  492/^  13.    25a;^  — 41a;y +I62/*. 

7.  25a^  — 9^2^2  +  16^^  14.    81a;*  — 34a;y +  1/*. 

*If,  in  Example  12,  Qy*  =(—  ^y^,  then  25 a;2?/2 should  be  added  to 
4x*  —  37  x2y2  +  9  y*,  in  order  to  make  the  expression  a  perfect  square. 
That  is,  we  should  have  : 

(4  X*  —  12  X2y2  +  9  y4)  _  25  x22/2, 
=  (2x2-3y2)2-25x2y2, 
=  (2x2- 32/2 +  5xy)  (2x2 -3?/2-5xy), 
or,  (2x2  +  5xy  -  3i/2)  (2x2  -  5x?/ -  3y2). 


FACTORS.  81 

137.   Case  XIV.     To  find  the  factors  of 

It  is  evident  that  the  first  terms  of  the  two  factors 
might  he  6x  and  x,  or  2x  and  3x,  since  the  product  of 
either  of  these  pairs  is  6a;^. 

Likewise,  the  last  terms  of  the  two  factors  might  be  12 
and  1,  6  and  2,  or  4  and  3  (if  we  disregard  the  signs). 

From  these  it  is  necessary  to  select  such  as  will  produce 
the  middle  term  of  the  trinomial.  And  they  are  found  by 
trial  to  be  3  a;  and  2x,  and  —  4  and  -f  3. 

,',6x'-i-x-12  =  (3x-4.)(2x  +  S). 

Exercise  43. 

Resolve  into  factors : 

1.  12x^  —  ox  —  2.  13.  6a^x^-\-ax  —  l. 

2.  12:r2-7ir  +  l.  14.  6b^-7bx-SxK 

3.  12a-2  — ^  — 1.  15.  4:X^^Sx-\-3. 

4.  3a;2  — 2ic  — 5.  16.  a^  —  ax  —  6x\ 

5.  Sx^-\-Ax~i.  17.  Sa^-\-Uab  —  15b^. 

6.  6x2  +  5.i-  — 4.  18.  (ja^  —  19ac-\-10c\ 

7.  4^2+13.^  +  3.  19.  Sx'-\-34:xy-\- 21  if. 

8.  4x2  +  ll.r-3.  20.  Sx'-22xi/-21i/, 

9.  4rX^  —  Ax  —  S.  21.  6x^  +  19x2/  — 7f. 

10.  x^  —  3ax-i-2a^.  22.  lla^  —  23ab-^2b\ 

11.  12a^  +  a2a:2  — x^  23.  2c^  —  13cd+6d\ 

12.  2^2_^5a;y  +  2/.  24.  6/+7z/^-3^^ 


82  ALGEBRA. 

138.  Case  XV.  The  factors,  if  any  exist,  of  a  polyno- 
mial of  more  than  three  terms  can  often  be  found  by  the 
application  of  principles  already  explained.  Thus  it  is 
seen  at  a  glance  that  the  expression 

fulfills,  both  in  respect  to  exponents  and  coefficients,  the 
laws  stated  in  §  83  for  writing  the  power  of  a  binomial ; 
and  it  is  known  at  once  that 

Again,  it  is  seen  that  the  expression 

consists  of  three  squares  and  three  double  products,  and 
from  §  79,  is  the  square  of  a  trinomial  which  has  for  terms 
X,  y,  z. 

It  is  also  seen  from  the  double  product  — 2xy,  that  x  and 
y  have  unlike  signs  ; 

and  from  the  double  product  2xz,  that  x  and  z  have  like 
signs.     Hence, 

x^  —  2xy-\-y^-^2xz  —  2yz-\-  z^=^(x  —  y-\- zf. 


Exercise  44. 
Resolve  into  factors : 

2.  a3  +  3a2  +  3a  +  l.  5.    ic*-4ic«+6a^-4rr  +  l. 

3.  a''  —  ^a^-\-Za  —  \.  6.    a''  —  4:a'c  +  6a^c^  —  4:ac^+c\ 

7.  x^-\-2xyi-f-{-2xz-]-27/z  +  z\ 

8.  x'  —  2xy-\-f  —  2xz-\-2yz-{-z\ 

9.  a'  +  b^-]-c'-{-2ab  —  2ac  —  2bc. 


FACTORS.  83 

139.  Case  XVI.     Multiply  2x  —  y-\-3hy  x-{-2y  —  S. 

2x-      7/  +  3 
X  +    2//  — 3 
2x'—    xy  +30! 

^xy  —  2if  -\-^y 

—6^  +  3;/  — 9 

2x'-\-2>xy  —  2if  —  ?>x-\-^y  —  'd 

It  is  to  be  observed  that  2x2  +  ^^y  —  2y2,  of  the  product,  is  obtained 
from  (2x  — ?/)X(x  +  22/); 

that  —  9  is  obtained  from  3  x  —  3 ; 

that  —  3x  is  the  sum  of  2x  X  —  3  and  x  x  3  ; 

that  ^y  is  the  sum  of  2y  X  3  and  —  y  x  —  3. 

From  this  result  may  be  deduced  a  method  of  resolving 
into  its  factors  a  polynomial  which  is  composed  of  two 
trinomial  factors.     Thus : 

Find  the  factors  of 

^x'  —  Ixy  —  Zif  —  9a! -f  30 ?/  —  27. 

The  factors  of  the  first  three  terms  are  (by  Case  XIV.) 
3x  +  y  and  2x  — 3y 

Now  —  27  must  be  resolved  into  two  factors  such  that  the  sum  of 
the  products  obtained  by  multiplying  one  of  these  factors  by  3x  and 
the  other  by  2x  shall  be  —  9x. 

These  two  factors  evidently  are  —  9  and  +  3. 

That  is,  (Q,x^-lxy-3f  —  ^x^my-21) 

=  (3a!  +  2/-9)(2^-32/+3). 

140.  The  following  method  is  often  most  convenient  for 
separating  a  polynomial  into  its  factors  : 

Find  the  factors  of 

2x^-^xy^2f-\-lxz-^yz-\'3zK 

1.  Keject  the  terms  that  contain  z. 

2.  Reject  the  terms  that  contain  y. 

3.  Reject  the  terms  that  contain  x. 


84  ALGEBRA. 

Factor  the  expression  that  remains  in  each  case. 

1.  2x'^  —  bxy  +  2>j^  =  {x  —  2y){2x  —  y). 

2.  2ic2  +  7xz  +  3z2  =  (x  +  32)  (2x  +  2). 

3.  2?/2-5?/z  +  3z2=(_22/  +  32;)(-2/  +  z). 

Arrange  these  three  pairs  of  factors  in  two  rows  of  three  factors 
each,  so  that  any  two  factors  of  each  row  may  have  a  common  term. 
Thus: 

x  —  2y,  x  +  3z,  -2?/+  3z; 

2x  —  y,  2x  +  z,  —y  +  z. 

From  the  first  row,  select  the  terms  common  to  two  factors  for  one 
trinomial  factor : 

x  —  2y  +  Sz. 

From  the  second  row,  select  the  terjns  common  to  two  factors  for 
the  other  trinomial  factor  : 

2x  —  y  +  z. 

Then,  2x''  —  5xi/^27f -\-7xz  —  5i/z  +  3z^ 

=  (x-2,/  +  3z)(2x-,j-j-z). 

141.  When  a  factor  obtained  from  the  first  three  terms 
is  also  a  factor  of  the  remaining  terms,  the  expression  is 
easily  resolved.     Thus  : 

x^  —  3xi/-{-27/  —  3x-\-6y, 
=  (x-2y)(x-y)-3{x-2y), 
=  (x  —  2y)(x  —  y  —  3). 

Exercise  45. 

Kesolve  into  factors  : 

1.  2x'  —  5xy-^2y'  —  17x-}-13y-}-21. 

2.  6x^  —  37xy-{-6f  —  5x  —  5y  —  l. 

3.  6x^  —  5xy  —  6y^  —  x  —  5y  —  1. 

4.  5x'  —  Sxy-}-3f-{-7x-5y-\-2. 

5.  2x^  —  xy  —  3y^  —  Sx-{-7y  +  6. 


FACTORS.  S5 

6.  x'  —  25i/-10x-20y-{-21, 

7.  2x^  —  5xy-]-2y^  —  xz  —  yz  —  z^. 

8.  6x^-\-xy  —  f  —  3xz-\-6yz  —  9^.. 

9.  6x'—7xy-\-y'-{'35xz  —  5yz  —  6^. 

10.  5x''  —  Sxy-^Sf  —  3xz-{-yz  —  2^, 

11.  2x'  —  xy  —  Sy^—5yz  —  2z^ 

12.  6x^  —  13xy  +  (jy'-^12xz  —  13yz-\-6zK  ■     - 

13.  x^  —  2 xy -\- y- -\- 5 X  —  5y. 

14.  2x^-\-5xy  —  3y^  —  4:Xz-\-2yz. 

Exercise  46. 

miscellaneous  examples. 

The  following  expressions  are  to  be  resolved  into  fac- 
tors by  the  principles  already  explained.  The  student 
should  first  carefully  remove  all  monomial  factors  from 
the  expressions. 

1.  5a!2— 15a;  — 20.  9       ^4_^^2_|_l 

2.  2x^  —  16x*-{-24:a^.  10.  x^  —  2/ —  xz -\- yz. 

3.  3a^b^~9ab  —  12.  11.  ab  —  ac  —  b^ -{- be, 

4.  a^-\-2ax-{-x^-{-4:a-{-4.x.  12.  3x^  —  3xz  —  xy-{-yz, 

5.  a^  —  2ab-\-b'  —  c\  13.  a^~x^  —  ab  —  bx. 

6.  x'—2xy-^y^  —  c^-\-2cd  —  d^l4.  a^  —  2ax-\-x^-j-a  —  x, 
>.  4  — x^  — 2x*^=^5':        '              15.  3x^  —  3y'  —  2x-{-2y. 

17.  aV  — aV-aV  +  1. 

18.  3a^-'2x^y  —  27xy^-{-lSy^, 


86;  ALGEBRA. 

19.  Ax^—x^+2x—l.     28.    4:a^  —  4:ah  +  b^ 

20.  x^  —  y^  29.    16x^  —  S0xy-]-100f. 

21.  x^-^-y^        .  30.    ^Qa^xhf  —  2Wxhf. 

22.  729  — x«.  31.    9ccy  — 30cc2/2^  +  25;s2^ 

23.  x^'^y-\-y^\  32.    16:z;^  — a;. 

24.  a*c  — c^  33.    x''  —  2xy  —  2xz-^7/-{-2yz-{-z\ 

25.  a;2  +  4:c  — 21.  34.    a^  —  ah  —  U^  —  4.a-\-12h. 

26.  3^2— 21aZ>+30^2^    35.    ic^  _^  2 a:?/ +  y^  _  a:  —  ^z  —  6. 

27.  2a^*— 4x'V— 6a^y.    36.    (a  +  ^)^  — c^ 

37.  x^  —  xy  —  Qif  —  4.x-{-12y.      39.    3x2  —  11x^  +  62/2. 

38.  1  — x  +  a;2  — x«.  40.    x2  +  20x  +  91. 

41.    (x  —  y){x''  —  z')  —  (x-z){x^  —  if). 

42.  ^2  — 5x  — 24.  50.    ?/2_4y_ii7, 

43.  (x2  — 2/2  — ;5;2^^2_4y2^2^  51^    ^2_^(3^_-^35_ 

44.  5 xy  + 5x22/^—60x^2^  52.    A:a^— 12 ah-[-W-4.c'. 

45.  3x«  — x2  +  3x  — 1.  53.    ((^  +  3^)2- 9  (^>  — c)2. 

46.  x2  —  2??ix  +  m2  —  n^.  54.    9x2 — ^y'^-\-4,yz  —  z^, 

47.  4a2^»2—  (a2+^,2_^2^2^  55^      ej2^2_  JJ^3_3^4^ 

48.  a'^a\  56.    a^- J^  — 3«7i  («  — Z*). 

49.  1  — 14a3x  +  49A2.  57.    x^^f-\-^xy(x-\-y). 

58.    a^  —  h^  —  a(a''-'h'')-\-h{a  —  h)\ 

59.  9x2^/2 _3^y3_g^4^  6Q_    6x2  +  13x2/  +  62/^. 

61.  ^a%''  —  ab^  —  12b\ 

62.  a2^2a6^  +  c^2_4^2_^;1^2^'c  — 9c2. 

63.  x^  — 2x22/  +  4x2/^  — 82/«.  64.    4 a2ic2  —  g a&x  +  3 ^»2. 


FACTORS.  87 

65.  lSx^  —  24:Xy  +  Sf-\-9x  —  6i/.  74.  16a^x  —  2x*. 

66.  2x'  +  2xy  —  122/-{-6xz-^lSyz.  75.  32ba^  —  Uf. 

67.  (x  +  7/y-l-xy(x-\-y  +  l).  76.  x-27x\ 

68.  x^  —  f  —  z^-\-2yz  +  x  +  y  —  z.  77.  x^  —  i/^ 

69.  2ic2_|.4^^_|_22^_^2aic  +  2ay.  78.  ^97n'  —  121n\ 

70.  16a^b  +  32abc-^12b(^.  79.  16  — 81 2/*. 

71.  m^p  —  rn^q  —  n^p -{- n^q.  80.  12^*  — ^2_g^ 

72.  12aa;2  — Uaxy  — 6a/.  81.  x^  — x^  +  a;  — 1. 

73.  2ic3  +  4a^2_7Q^  g2^  ^2_j_2^4_l_y2, 

83.    49(a  — ^>)2  — 64(m  — 7i)l 

85.  ic2_53^._^350. 

86.  ic^  — 2arv/4-x2  — 4cc4-87/  — 4. 

87.  2ai  — 2^»c  — ae  +  ce  +  2i2  — ie. 

88.  125a:*  +  350a;y  +  245a;2/^ 

89.  a^^a''b  +  a'b^^a%^^a'b^  +  al/', 

90.  2a*x  — 2a^ca;  +  2ac^a;--2c*x. 

91.  6x2  — 5xy  — 6y*  +  3a;;s;  +  15r/;sj  — 9«2. 

92.  4a;2  — 9x^  +  22/^-30;^— y«  —  «2^ 

93.  3^2  — 7a^  +  2^»2_|_5^^_5jg_|_2c2. 

94.  x^  — 2x^  +  0:2  — 8x  +  8. 

95.  5x2  — 8x?/  +  32/--5x  +  32/. 

96.  a''  —  2ad-\-d^  —  U^^12bc  —  9e', 

97.  (x2  — X  — 6)(x2  — X  — 20). 


CHAPTER  VII. 
Common  Factors  and  Multiples. 

142.  A  common  factor  of  two  or  more  expressions  is  an 
expression  which  is  contained  in  each  of  them  without  a 
remainder.     Thus, 

5  a  is  a  common  factor  of  20  a  and  25  a; 
3iry  is  a  common  factor  of  12a;y  and  15a;y. 

143.  Two  expressions  which  have  no  common  factor 
except  1,  are  said  to  be  prime  to  each  other. 

144.  The  Highest  Common  Factor  of  two  or  more  ex- 
pressions is  the  product  of  all  the  factors  common  to  the 
expressions. 

Thus,  3a^  is  the  highest  common  factor  of  3a^,  ^a^,  and 
12  a^ 

5xhf  is  the  highest  common  factor  of  10 «y  and  15 ^y. 

For  brevity,  H.  C.  F.  will  be  used  for  Highest  Common 
Factor. 

(1)  Find  the  H.  C.  F.  of  42  a^^^^  and  21  a''h^x\ 

4.2a'b'x  =2X3X7Xa'XPXx', 
21a^b^x^  =  3  XTXa^Xh^Xx^ 
.'.  the  H.  C.¥.=3  X  7  X  a^  X  P  X  X 
=  21a^'x. 

(2)  Find  the  H.  C.  F.  of  2  a^a^  +  2  ax""  and  3  abxy  +  3  bx^y. 

2  a^x  +  2  ax^  =  2  ax  (a -]- x) ; 
3  abxy  +  3  bx^y  =  3  bxy  (a-\-x). 
.-.theH.  C.  F.==x(a  +  a;). 


COMMON    FACTORS    AND    MULTIPLES.  89 

(3)  Find  the  H.  C.  F.  of 

8  aV  —  24  a^x  4- 16  a^  and  12  ax^y  — 12  axy  —  24  ay. 

Sa^x^  —  24.a'x  +  16a^  =  Sa\x'-3x  +  2), 
=  2V(x  — l)(ic  — 2)j 
12  axV  — 12  ax^  —  24  ay  =  12  ay  (x^  —  a;  —  2), 

=  22x3ay(a;  +  l)(a;  — 2). 
/.theH.  C.  r.=22a(x-2), 
=  4a(a:  — 2). 

Hence,  to  find  the  H.  C.  F.  of  two  or  more  expressions  : 
Resolve  each  expression  into  its  lowest  factors. 
Select  from  these  the  lowest  power  of  each  common  factory 
and  find  the  product  of  these  powers. 

Exercise  47. 
Find  the  H.  C.  F.  of  : 
1.   l^ab^c'd  and  ^Qa^'bcdK  2.   VJpq^y  34 ^V;  and  51py. 

3.  8a^/«*,  Vlo^f^y  and  20xVV. 

4.  30icV,  90  a; V,  and  120  a^y*. 

5.  a^  —  h^  and  a"  —  h^.  7.    a'  +  a»  and  (a  +  ic)«. 

6.  a^  — ic^and  (a  — a:/.  8.    Qx^  — 1  and  (3a;-f  1)». 

9.  7x*~4a;  and  7a*a;  — 4a* 

10.  12aVy  —  4a8V  and  30 a Vy^  — 10 aVy». 

11.  8a%  — 12 a2^»c''  and  6aZ>^c  +  4aJV. 

12.  a;2__2a;  — 3  and  a^  +  x  —  12. 

13.  2a3-2a^.2and45(a  +  ^»)l 

14.  12x^2/ (x  —  i/)(x  —  3y)  andl8x2(a;  — y)(3x  — y). 

15.  3x3+6x2  — 24a;  and  6x«  —  96x. 


^  ALGEBRA. 

16.  ac  (a  —  h)(a  —  c)  and  he  (h  —  a)(b'—  c). 

17.  lOx^j  —  60  xV  -j-  5  xif  and  5  xy  —  5xf  — 100 y\ 

18.  x(x  +  iy,  x\x^  —  l),  and  2x(x^  —  x  —  2). 

19.  3ic2_(3^_|_3^  62c2  +  6:k  — 12,  and  12x2  —  12. 

20.  6  (6^  -  ^>)^  8  (a'  -  ly,  and  10  (a^  —  h'). 

21.  a;2  — y2^  (a;  +  2/)2,  and£c2  +  3ic?/  +  2?/2. 

22.  x^  —  'if,  x^  —  if,  and  a;^—  lxy-\-Q>y\ 

23.  a;2  —  1,  x^  —  1,  and  a;^  +  x  —  2. 

145.  When  it  is  required  to  find  the  H.  C.  F.  of  two  or 
more  expressions  which  cannot  readily  be  resolved  into 
their  factors,  the  method  to  be  employed  is  similar  to  that 
of  the  corresponding  case  in  arithmetic.  And  as  that 
method  consists  in  obtaining  pairs  of  continually  decreas- 
ing numbers  which  contain  as  a  factor  the  H.  C.  F.  required; 
so  in  algebra,  pairs  of  expressions  of  continually  decreasing 
degrees  are  obtained,  which  contain  as  a  factor  the  H.  C.  F. 
required. 

The  method  depends  upon  two  principles : 

1.  Any  factor  of  an  expression  is  a  factor  also  of  any 
multiple  of  that  expression. 

Thus,  if  F  represent  a  factor  of  an  expression  A ,  so  that  A  =  nF, 
then  mA  =  mnF.     That  is,  mA  contains  the  factor  F. 

2.  Any  common  factor  of  two  expressions  is  a  factor  of  the 
sum  or  difference  of  any  multiples  of  the  expressions. 

Thus,  if  F  represent  a  common  factor  of  the  expressions  A  and  B 
so  that 

A  =  mF,  and  B=  nF; 
then  pA  =  pmF,  and  qB  =  qnF. 

Hence,        pA±qB  =  pmF  ±  qnF, 
=  {pm  ±  qn)F. 
That  is,  pA  ±qB  contains  the  factor  F. 


COMMON    FACTORS    AND    MULTIPLES.  I>t 

146.   The  general  proof  of  this  method  as  applied  to 
numbers  is  as  follows : 

Let  a  and  b  be  two  numbers,  of  which  a  is  the  greater. 
The  operation  may  be  represented  by : 


b)a{p 

42)154(3 

nF)mF(p 

pb 

126 

pnF 

c)b(q 

28)42(1 

cF)nF{q 

qc 

28 

qcF 

d)c(r 

14)28(2 

F)cF(c 

rd 

28 

cF 

p,  q,  and  r  represent  the  several  quotients, 
c  and  d  represent  the  remainders, 
and  d  is  supposed  to  be  contained  exactly  in  c. 
The  numbers  represented  are  all  integral. 

Then  c  =  rd. 


b=  qc  +  d=  qrd+  d=  {qr-\-l)d, 
a  =  pb  +  c  =  pqrd  +  pd  +  rd, 
=  {pqr  +  p  +  r)d. 


.'.  dis  a  common  factor  of  a  and  h. 

It  remains  to  show  that  d  is  the  highest  common  factor  of  a  and  b. 

Let  /  represent  the  highest  common  factor  of  a  and  b. 

Now  c=  a  —  pby  and  /  is  a  common  factor  of  a  and  6. 

.-.  by  (2)  /  is  a  factor  of  c. 

Also,  d  =  6  —  5C,  and  /  is  a  common  factor  of  b  and  c. 

.-.  by  (2)  /  is  a  factor  of  d. 

That  is.  d  contains  the  highest  coriinon  factor  of  a  and  b. 

But  It  has  been  shown  tiiat  d  is  a  common  factor  of  a  and  6. 

.-.  d  is  the  highest  common  factor  of  a  and  6. 

Note.  The  second  operation  represents  the  application  of  the 
method  to  a  particular  case.  The  third  operation  is  intended  to  rep- 
resent clearly  that  every  remainder  in  the  course  of  the  operation 
contains  as  a  factor  the  H.  C.  F.  sought,  and  that  this  is  the  highest 
factor  common  to  that  remainder  and  the  preceding  divisor. 


9Sl  ALGEBRA. 

147.  By  the  same  method,  find  the  H.  C.  F.  of 

2a^  +  x  —  S)4:X^-\-Sx^—    x  —  6{2x-{-S 
4:x^-i-2x^  —  6x 

6x^-\-5x  —  6 
6a;'  +  3a;  — 9 

2x-{-3)2x'+    x  —  S(x-l 
2y?^Zx 
—  2x  —  S 
/.theH.  F.  C.  =  2a;  +  3.  -2a;-3 

The  given  expressions  are  arranged  according  to  the  descending 
powers  of  x. 

The  expression  whose  first  term  is  of  the  lower  degree  is  taken  for 
the  divisor  ;  and  each  division  is  continued  until  the  first  term  of  the 
remainder  is  of  lower  degree  than  that  of  the  divisor. 

148.  This  method  is  of  use  only  to  determine  the  com- 
pound factor  of  the  H.  C.  F.  Simple  factors  of  the  given 
expressions  must  first  be  separated  from  them,  and  the 
highest  common  factor  of  these  must  be  reserved  to  be 
multiplied  into  the  compound  factor  obtained. 

Find  the  H.  C.  F.  of 

12x*  +  30x^  —  72x'a,udS2x^-\-S4x^-lT6x, 

12x*  +  30x^  —  72x^  =  6x'(2x^  +  5x  —  12). 
32x3  +  84x2  — 176x:-=4cc  (8x2  + 21a;  — 44). 
6a^  and  4x  have  2x  common. 

2a^  +  5x  — 12)8x2  +  21x-44(4 
8x2  +  20x-48 

x+   4)2x2  +  5x  — 12(2x  — 3 
2x2  +  8x 

—  3x  — 12 
.-.theH.  F.  C.  =  2x(x  +  4).  — 3x  — 12 


COMMON    FACTORS    AND    MULTIPLES.  93 

149.    Modifications  of  this  method  are  sometimes  needed. 

(1)  Find  the  H.  C.  F.  of  4x-  —  Sx  —  d  and  12x-  —  Ax  —  65. 

4^2  — 8x  — 5)12a:2—   4a;  — 65(3 
12a;^  — 24a;  — 15 
20x  — 50 

The  first  division  ends  here,  for  20  x  is  of  lower  degree  than  4  x'^. 
But  if  20  a;  — 50  be  made  the  divisor,  4x2  will  not  contain  20  x  an 
integral  number  of  times. 

Now,  it  is  to  be  remembered  that  the  H.  C.  F.  sought  is  contained 
in  the  remainder  20  x  —  50,  and  that  it  is  a  compound  factor.  Hence 
if  the  simple  factor  10  be  removed,  the  H.  C.  F.  must  still  be  con- 
tained in  2  X  —  5,  and  therefore  the  process  may  be  continued  with 
2  X  —  5  for  a  divisor. 

2x-5)4:x'-    8a;-5(2a;  +  l 
4a:^-10x 

2a;-5 
2  a;  — 5 

.-.theH.  C.  F.  =2a;  — 5. 

(2)  Find  the  H.  C.  F.  of 

21a;3  — 4a;2  — 15a;-2and21a;3-32a;2-54a;-7. 

21a;«-4a-='-15a;-2)21a;^-32x2  — 54a;-7(l 
21x^—   4a;^-15a;-2 
-28a-2  — 39a;  — 5 

The  difficulty  here  cannot  be  obviated  by  removing  a  simple  factor 
from  the  remainder,  for  —  28  x^  —  39  x  —  5  has  no  simple  factor.  In 
this  case,  the  expression  21  x^  —  4  x^  —  15  x  —  2  must  be  multiplied  by 
the  simple  factor  4  to  make  its  first  term  divisible  by  —  28  x^. 

The  introduction  of  such  a  factor  can  in  no  way  affect  the  H.  C.  F. 
sought ;  for  the  H.  C.  F.  contains  only  factors  common  to  the  remain- 
der and  the  last  divisor,  and  4  is  not  a  factor  of  the  remainder. 

The  signs  of  all  the  terms  of  the  remainder  may  be  changed  ;  for 
if  an  expression  A  is  divisible  by  —  F,  it  is  divisible  by  +F. 


^j|  ALGEBRA. 

The  process  then  is  continued  by  changing  the  signs  of  the  remain- 
der and  multiplying  the  divisor  by  4. 

2Sx^-{-S9x-i-5)S4:X^—   16x'—   60x—   S(Sx, 
S4.x^  +  117x^-\-   15x 

—  133rz;2—    T5x—   8 
Multiply  by  —  4,  --4 


532  0^2  _^  300  a; +  32  (19 
532^^  + 741a; +  95 
Divide  by  -  63,  —  63)— 441(^  —  63 

Ix-i-    1 

7a;  +  l)28£c2  +  39a;  +  5(4a;  +  5 
28a:^+   Ax 

350^  +  5 
.-.  the  H.  C.  r.  =  7a:  +  l.     35a; +  5 

(3)  Find  the  H.  C.  F.  of 

8a;2  +  2a;  — 3and6a;3+5ic2  — 2. 

6a;^+    5x'—   2 
Multiply  by  4,       4 


8a;2  +  2a^-3) 

^24:X 

^  +  20a;2—    8    (3a;  +  7 

24:  X 

3+    6x'—   9x 

14x2+    9^_   8 

Multiply  by  4, 

4 

56a;2  +  36ic-32 
56a;2+14a;  — 21 

Divide  by  11, 

11)  22  a; -11 

2x—   l)8a;'  +  2a;  — 3(4aj  +  3 
8a;2-4a; 

6a;-3 

.-.theH.  a 

.¥.= 

=  2a;-l.                                 6x  —  ^ 

In  this  case  it  is  necessary  to  multiply  by  4  the  given  expression 
6cc3  +  5x2  —  2  to  make  its  first  term  divisible  by  Sx^,  4  being  obvi- 
ously not  a  common  factor. 


COMMON    FACTORS    AND    MULTIPLES. 


95 


The  following  arrangement  of  the  work  will  be  found 


most  convenient : 

Sx'  +  2x-S 

6x'+   5x'-  2 

Sx^-4.x 

4 

6x  —  3 

24.x'-{-20x^-   8 

3x 

6x-S 

24a;8+    6x'-   9x 

Ux'-j-   9x—   8 

4 

56x'  +  36x-32 

+  7 

56x^  +  Ux-21 

ll)22aj  — 11 

2x-   1 

4.X  +  3 

150.  From  the  foregoing  examples  it  will  be  seen  that, 
in  the  algebraic  process  of  finding  the  highest  common 
factor,  the  following  steps,  in  the  order  here  given,  must 
be  carefully  observed : 

I.  Simple  factors  of  the  given  expressions  are  to  be  re- 
moved from  them,  and  the  highest  common  factor  of  these 
is  to  be  reserved  as  a  factor  of  the  H.  C.  F.  sought. 

II.  The  resulting  compound  expressions  are  to  be  ar- 
ranged according  to  the  descending  powers  of  a  common 
letter  ;  and  that  expression  which  is  of  the  lower  degree  is 
to  be  taken  for  the  divisor  ;  or,  if  both  are  of  the  same 
degree,  that  whose  first  term  has  the  smaller  coefficient. 

III.  Each  division  is  to  be  continued  until  the  remainder 
is  of  lower  degree  than  the  divisor. 

IV.  If  the  final  remainder  of  any  division  is  found  to 
contain  a  factor  that  is  not  a  common  factor  of  the  given 
expressions,  this  factor  is  to  be  removed  ;  and  the  resulting 
expression  is  to  be  used  as  the  next  divisor. 

y.  A  dividend  whose  first  term  is  not  exactly  divisible 
by  the  first  term  of  the  divisor,  is  to  be  multiplied  by  such 
an  expression  as  will  make  it  thus  divisible. 


96 


ALGEBRA. 


Exercise  48. 
Find  the  H.  C.  F.  of  : 

1.  5x'+4:x  —  l,  20x'  +  21x  —  5. 

2.  2x^-'4.x'—lSx  —  7,  6£c3--lla^2_3j^_20. 

3.  6a^+25a^  —  21a^  +  4:a,  24.a* +  112  a^  — 94:  a^  +  IS  a. 

4.  9x^-{-9x^  —  4.x  —  4:,  A5x^  +  54:X^  —  20x  —  24:. 

5.  2Tx^  —  3x^-{-6x^  —  3x',  162a;«  +  48a;3  — 18a;2  +  6a;. 

6.  20x^  —  60x^-{-50x-20,  32x'-92x'  +  6Sx^-24.x. 

7.  4:X^  —  Sx  —  5,  12a;2  —  4x  — 65. 

8.  Sa^  —  5a^x  —  2ax%  9a^  —  Sa^x  —  20ax^ 

9.  10x^  +  x^  —  9x  +  24:,  20a:*  — 17^2  + 48a;  — 3. 

10.  8a;«  — 4x2  — 32a:  — 182,  sea:^  — 84a;2  — Ilia;  — 126. 

11.  5a;2(12a;3+4a;2+17a;— 3),  10a;(24a;3— 52a;2+14a;-l). 

12.  9x*i/  —  xy  —  20xy*,  18x^y  —  lSxy  —  2xf  —  Sy\ 

13.  6a;2  — a;  — 15,  9a;2  — 3a;  — 20. 

14.  12a;«— 9a;2  +  5a;  +  2,  24a;2+10a;  +  l. 

15.  6a;^  +  15a;2  — 6a;  +  9,  9a;3+6a;2  — 51a;  +  36. 

16.  4.x^-x'y  —  xif-5f,  7x^-}'4.x'y  +  4.xf  —  3f. 

17.  2a^  —  2a^  —  3a  —  2,  3a^  —  a'^  —  2a  —  16. 

18.  122/3  +  22/2  —  942/  — 60,  482/'  — 242/2_343^_|_3q^ 

19.  9a;(2a;*  — 6a;2  — a;2+15a;  — 10), 

6a;2  (4a;*  + 6a;3  —  4a;2  — 15a;  — 15). 

20.  15x^+2x^—75x'-}-5x  +  2,  35a;*H-a;S-175a;2+30a;+l. 

21.  21a;*  — 4a;3  — 15a;2  — 2a;,  21a;«  — 32a;2  — 54a;  — 7. 

22.  9a;V— 22a;y— 3a;2/*+102/^  9a;V - 6 a;y+a;y— 25 a;2/«. 


COMMON    FACTORS    AND    MULTIPLES.  97 

23.  6a^  —  4:X*-llx^  —  3x^  —  3x  —  l, 

4:x'-i-2x'-lSx'-\-3x  —  5. 

24.  x*  —  ax^  —  a^x^  —  a^x  —  2a\  3x^  —  7ax^-{-3a^x  —  2a^ 

161.  The  H.  C.  F.  of  three  expressions  will  be  obtained 
by  finding  the  H.  C.  F.  of  two  of  them,  and  then  of  that 
and  the  third  expression. 

For,  if  A,  B,  and  C  are  three  expressions, 

and  D  the  highest  common  factor  of  A  and  B, 
and  £J  the  highest  common  factor  of  Z>  and  C, 
then  D  contains  every  factor  common  to  A  and  B, 

and  ^  contains  every  factor  common  to  D  and  C. 
.'.  E  contains  every  factor  common  to  A,  B,  and  C. 

Exercise  49. 
Find  the  H.  C.  F.  of : 

1.  2x^-\-x  —  l,  0^2-1-5:^4-4,  a-3-f  1. 

2.  f-f-^j  +  1,  3f-2i/-l,  i/-?/-\-,j-l. 

3.  x^-4.x'-{-9x-10,  x^-\-2x'-3x-{-20,  0:^+5 3:^-9 a; +35. 

4.  0^3-7:^2  + 16a:  — 12,  Sa;^  — Ua-^  +  lGa;, 

5x^  —  10x'-\-7x  —  U. 

5.  f-5if+ny-15,  ,/-,/  +  3y-\-5, 

2y'-7f-{-16y-15. 

6.  2x^-\-3x  —  5,  3x^  —  x  —  2,  2x^-\-x  —  3. 

7.  x^  —  1,  x^  —  x^—x  —  2,  2x^  —  x^  —  x  —  3. 

8.  x^  — 3x  — 2,  2x^  +  3x2  —  1,  a;3  +  l. 

9.  12(x'-y%  10(x'-f),  8(A  +  V> 

10.  x*-\-xy^,  xhj-\ry\  x^-\-xhf-\-if. 

11.  2{7?y-xf),  3{xhj-xf),  ^{xHj-xy%  5(x^y-xf). 


ALGEBRA. 


Lowest  Common  Multiple. 


162.  A  common  multiple  of  two  or  more  expressions  is 
an  expression  which  is  exactly  divisible  by  each  of  them. 

153.  The  Lowest  Common  Multiple  of  two  or  more  ex- 
pressions is  the  product  of  all  the  factors  of  the  expres- 
sions, each  factor  being  written  with  its  highest  exponent. 

154.  The  lowest  common  multiple  of  two  expressions 
which  have  no  common  factor  will  be  their  product. 

For  brevity  L.  C.  M.  will  be  used  for  Lowest  Common 
Multiple. 

(1)  Find  the  L.  C.  M.  of  12 a%  Ubc^,  S6ab\ 

12ah  =  2''X3a\ 
Ubc^  =  2  XTbc", 
36ab^  =  2^X3^aP. 

.♦.  The  L.  C.  M.  =2^  X  3^  X  1  a^b^c'  =  262a%^(^. 

(2)  rind  the  L.  C.  M.  of 

2a^-{-2ax,  6a^  —  6x^,  Sa^  —  6ax-\-Sa^. 

2a?-\-2ax  =2a(a-{-x), 

6a^  —  6x^  =2X3(a  +  x)  (a  —  x), 

3a'^  —  Q>ax-\-3x}  =  3{a  —  x)\ 

.-.the  L.  C.  M.  =  6  a  {a  +  ic)  (a  —  x)\ 

Exercise  50. 
Find  the  L.  C.  M.  of 

1.  4 A,  6aV,  2ax\  4.    x^—1,  x?  —  x, 

2.  18  oa^  72  ^2/^  12  xy,  5.    a^  —  b^,  a^-^ab. 

3.  x",  ax-^-x".  6.    2x  —  l,  4.x''  — 1. 


COMMON    FACTORS    AND    MULTIPLES.  99 

7.  a  +  b,  a'  +  b^  9.    x'-x,  x'-l,  x^  +  1. 

8.  x^-1,  x'-\-l,  x'-l.  10.    x'-l,  x'-x,  x'-l. 

11.  2a  +  l,  4^2-1,  Sa^  +  l. 

12.  (a  +  by,  a:'  —  b\ 

13.  4:(l-\-x),  4(1-0;),  2(l-x^. 

14.  x  —  1,  x^  +  x-}-!,  x^  —  1. 

15.  x'-f,   (x  +  yy,   (x-yy. 

16.  x'-f,  3  (x-yy,  12(x'  +  f). 

17.  6(x'  +  xy),  S(xy-f),  10(x'-f), 

18.  x^-{-5x-]-(j,  x'-}-6x  +  8. 

19.  a^  —  a  —  20,  a^-\-a  — 12. 

20.  a;2  4.iia;  +  30,  x' -{- 12 x -\- 35. 

21.  a:2_9a;_22,  a;2  — 13a:  +  22. 

22.  4ai(a2_3^^,_}_2^,2^)^  5a2(^2_j_^^_gj2^^ 

23.  20(a;2-l),  24(a;2-a;~2),  16{x'-{-x-2). 

24.  12a:y(a:2-2r0,  2a:2(x  +  yn  32/^ (a- -y)^. 

25.  (a-b)(b-c),   (b-c)(c  —  a),   (c-a)(a  —  b). 

26.  (a  —  b)(a  —  c),   (b  —  a)(b  —  c),   (c  —  a){c  —  b). 

27.  x3-4x2  +  3a^,  x^-{-x^-12x\  x'  +  3x'-4.x\ 

28.  xV-a^y^,  3x(a;-y)2,  Ay(x-yy. 

29.  (a  +  Z»)2-(c  +  c/)^  (a  +  c)2-(^»  +  ^^  («  +  ^)2_(^ 4.^)2. 

30.  (2a;— 4)(3a;  — 6),  (ic  — 3)(4a;  — 8),  (2ic  — 6)(5a;  — 10). 


155.  When  the  expressions  cannot  be  readily  resolved 
into  their  factors,  the  expressions  may  be  resolved  by  find- 
ing their  H.  C.  F. 


100  ALGEBBA. 

Find  the  L.  C.  M.  of 


6x^—11x^1/  ■\-2f 


3x'y  +  4.xf-\-22f 
Sxhj±Axf±2f 


9x^-22xf-Sy^ 
2 


lSx^  —  Uxif  —  16y^ 
lSx^  —  3Sxhj-\-   6f 


lly)33x^y-Uxf  —  22f 


2x—y 


3x^    —   4.xy  —   2f 

Hence,  6x^  —  llx^y  +  27/=(2x  —  y)(3x^  —  4.xy  —  27/), 
and  9x^  —  22xy'  —  Sy^=(3x-\-4:y)(3x^  —  4.xy  —  2f). 

.-.the  L.  C.  M.  =(2x  —  y)(3x-i-4:ij)(3x''  —  4.xy  —  2f). 

In  this  example  we  find  the  H.  C.  E.  of  the  given  expres- 
sions, and  divide  each  of  them  by  the  H.  C.  F. 

156.  It  will  be  observed  that  the  product  of  the  H.  C.  F. 
and  the  L.  C.  M.  of  two  expressions  is  equal  to  the  product 
of  the  given  expressions.     For, 

Let  A  and  B  denote  the  two  expressions,  and  D  their 
H.  C.  F. 

Suppose  A  =  aD,  and  B  =  hD  \ 

Since  D  consists  of  all  the  factors  common  to  A  and  B, 
a  and  b  have  no  common  factor. 

.*.  L.  C.  M.  of  a  and  h  is  ah. 

Hence,  the  L.  C.  M.  oi  aD  and  bl>  is  abD: 

Now,  A  =  aD,  and  B  =  bD  j 

.\AB  =  abDXD. 

AB 

.'.  -J- =  abD  =  the  lowest  common  multiple.     That  is. 

The  L.  C.  M.  of  two  expressions  can  be  found  by  dividing 
their  product  by  their  H.  C.  F. 

Or,  by  dividing  one  of  the  expressions  by  the  H.  C.  F.,  and 
multiplying  the  result  by  the  other  expression. 


COMMON    FACTORS    AND    MULTIPLES.  101 

157.  To  find  the  L.  C.  M.  of  three  expressions  A^  B,  C. 
Find  if,  the  L.  C.  M.  of  ^  and  ^ ;  then  the  L.  C.  M.  of  M 
and  C  is  the  L.  C.  M.  required. 


Exercise  61. 
Find  the  L.  C.  M.  of  : 

1.  ^x^  —  x  —  2,  2\x^  —  Vlx'\-2,  14a;2  +  5a;  — 1. 

2.  x^-l,  x?-\-'Zx-Z,  6x^-x  —  2. 

3.  a:^  — 27,  x'  —  15x  +  S6,  x^  —  Sx'  —  2x-}-6. 

4.  5x'-\-19x-4:,  10a:2  +  13a;-3. 

5.  12x'-^ X1/-61/,  lSx^-\-lSxi/-20y'. 

6.  x*  —  2x^-]-x,  2a:<  — 2it;«  — 2a;  — 2. 

7.  12a;2  +  2x-4,  12  a:^- 42  a:  — 24,  12x2 -.28  a;  — 24. 

8.  x^-6x^-\-llx  —  6,  x^  — 9x2  +  26a;  — 24, 

a;«-8a;2-f  19a;-12. 

9.  x^  — 4a2,  x^-i-2ax^  +  4:a^x-{-Sa% 

x^  —  2a^-\'^a''x  —  Sa^ 

10.  0^  +  2x^1/  — xy^-'2y^,  x^  —  2x?y  —  x%^'\-2f, 

11.  l+j9+y,  l-i>+i>^  l+y+i>*. 

12.  (1-a),   il-a)\   i\-a)\ 

13.  (a  +  c)2-^»^   (a  +  ^»)2  — c2,   (h-^cf  —  a\ 

14.  3c3  — 3cV  +  c2/^  — 2/",  4c3  — cV  — 3cy2. 

15.  m^  — 8w  +  3,  ?/i«  +  3m*4-m  +  3. 

16.  207^*+7i2_l^    2571^+571^-71-1. 

17.  h^  —  2J?^y'  —  U^r'^,  4^»3— 12Z>2-|-9J  — 1. 

18.  2r«— 8r^+12r3  — 8r2  +  2r,  3^5_  6y.3_|.3^^ 


CHAPTEE  VIII. 

Fractions. 

4 

158.  In  Arithmetic,  an  expression  of  the  form  of  -  is 

employed  to  indicate  that  4  units  are  divided  into  5  equal 
parts,  and  that  one  of  these  parts  is  taken  ;  or,  that  one 
unit  is  divided  into  5  equal  parts,  and  that  4  of  these  parts 
are  taken. 

The  expression  is  also  used  to  indicate  the  quotient  of  ^ 
divided  by  5. 

4 

159.  The  expression  -  is  called  a  fraction.    4  is  the  nu- 

o 

merator,  and  5  the  denominator. 

160.  The  numerator  and  denominator  together  are  called 
the  terms  of  the  fraction. 

161.  The  denominator  shows  into  how  many  equal  parts 
the  unit  is  divided  and  therefore  names  the  part ;  and  the 
numerator  shows  the  number  of  the  parts  taken. 

It  will  be  observed  that  a  figure  written  above  the  line 
in  a  fraction  serves  a  very  different  purpose  from  that  of  a 
figure  written  below  the  line. 

A  figure  written  above  the  line  denotes  number ; 

A  figure  written  below  the  line  denotes  name. 

162.  Every  whole  number  may  be  written  in  the  form 

4 

of  a  fraction  with  unity  for  its  denominator  ;  thus,  4  =  -* 


FRACTIONS.  103 

To  Keduce  a  Fraction  to  its  Lowest  Terms. 

163.  Let  the  line  AB  be  divided  into  5  equal  parts,  at 
the  points  C,  D,  E,  F. 

^1    I    I    I    I    I    I    I    I    I    I    t    I    I    I    It? 

C  B  E  F 

Then  AF  is  f  of  AB.  (1) 

Now  let  each  of  the  parts  be  subdivided  into  3  equal  parts. 

Then  AB  contains  15  of  these  subdivisions,  and  AF  con- 
tains 12  of  these  subdivisions. 

.• ,  AF  i^  \l  oi  AB.  (2) 

Comparing  (1)  and  (2),  it  is  evident  that  |  =  -J-f. 

In  general : 

If  we  suppose  AB  to  be  divided  into  b  equal  parts,  and 
that  AF  contains  a  of  these  parts,  where  a  and  b  signify 
positive  whole  numbers j 

TYiQiiAFi^^oiAB.  (3) 

Now,  if  we  suppose  each  of  the  parts  to  be  subdivided 
into  c  equal  parts,  c  being  a  positive  whole  number. 

Then  AB  contains  be  of  these  subdivisions,  and  AF  con- 
tains ac  of  these  subdivisions. 

.'.AFisj-^oiAB.  (4) 

Comparing  (3)  and  (4),  it  is  evident  that 
a ac 

Since  —  is  obtained  by  multiplying  by  c  both  terms  of 

the  fraction  -> 
b 

and,  conversely,  -  is  obtained  by  dividing  by  c  both  terms 

of  the  fraction  -7-5  it  follows  that 
be 


104  ALGEBRA. 

If  the  numerator  and  denominator  of  a  fraction  he  multi- 
plied by  the  same  number,  or  divided  by  the  same  number, 
the  value  of  the  fraction  is  not  altered. 

In  Algebra,  a  fraction  simply  indicates  the  quotient  of 
the  numerator  divided  by  the  denominator,  but  the  laws 
that  apply  to  fractions  in  Arithmetic  apply  also  to  frac- 
tions in  Algebra. 

164.    To  reduce  a  fraction  to  lower  terms, 

Divide  the  numerator  and  denominator  by  any  common 

factor. 

A  fraction  is  expressed  in  its  lowest  terms  when  both 
numerator  and  denominator  are  divided  by  their  H.  C.  F. 

Reduce  the  following  fractions  to  their  lowest  terms  : 

a^  —  x^ (a  —  X)  (a^  -\-  ax  -{-  x^) a^-\-ax-\-  x^ 

^  ^  a^  —  x^  (a  —  x)(a-\-x)  a-\-x 


(2)      .2 


^^+7a  +  10_(a  +  5)(a  +  2)^a  +  5, 
a^  +  Sa  +  e  ~(a  +  3)(a  +  2)"~a  +  3* 


Qx^'  —  ^x  —  Q  _(2a;  — 3)(3x  +  2)_3a;-f2 
^^^  8x2— 2a;  — 15~(2x  — 3)(4x  +  5)""4aj  +  5 

a«-7a24-16^-12 
^^      Sa^  —  l'ia^^lQa   ' 

Since  in  Ex.  (4)  no  common  factor  can  be  determined 
by  inspection,  it  is  necessary  to  find  the  H.  C.  F.  of  the 
numerator  and  denominator  by  the  method  of  division. 


FRACTIONS. 


106 


Suppress  the  factor  a  of  the  denominator  and  proceed  to  divide : 


7a2  +  16a-    12 


3a« 
3a« 

-21a2  +  48a- 
—  14a2  +  16a 

36 

—   7a^  +  32a- 
3 

36 

—  21a2-|-96a- 
— 21a2-|.98a- 

108 
112 

-2)—   2a  + 

4 

14a4-16 


3a2—    6a 


a—      2 
.-.theH.  C.  F.  =  a-2. 

Now,  if  a«  — 7a2+16a  — 12  be  divided  by  a  — 2,  the 
result  is  a^  — 5a  +  6;  and  if  3a^  — 14a^+16a  be  divided 
by  a  —  2,  the  result  is  3  a^  —  8  a. 

.   <t'  — 7a'-|-16a-12_a^  — 5a4-6 
•'*    3a«  — 14a2+16a    ~    3a2  — 8a  * 

166.  If  the  common  factors  cannot  be  determined  by 
inspection,  the  H.  C.  F.  must  be  found  by  division. 


Exercise  b2. 
Reduce  to  lowest  terms : 


1. 


3. 


4. 


6. 


4ic(ir  +  l) 
a;^-9a;  +  20 
a^=^-7x  +  12* 
a;^~2a;-3 
ic*— lOx  +  21* 

ic^  +  x  +  l* 

x^'  —  f 


6. 


8. 


9. 


10. 


a«  +  l 


a«-f  2a24-2a  +  l 
a^  — a  — 20 


a2  +  a-12 

ic«  +  2a;=^  — 3x  +  20* 

x^—'6x^-\-\lx  —  \^ 

x^  —  7?-\rZx-\-h 

x^  —  x^y  —  x^  —  2/* 


106 


ALGEBBA. 


H*  Q         7i        i     !^*  21. 


a^  —  3a-\-2  a^  —  a?h  —  a?W^ 


3x^  +  2x-l  (a  +  by 

x^-i-x'  —  x  —  l  a?-ab-2h 


„     x^  —  Zx^-\-^x  —  2  Zab(a}  —  b'^ 

8a;«-27a«       *  a2_^a^>  — ac 

^  *    9a'+3a^-2&2'  ^^'     ^x'y  —  hxy'  —  ^f 

ic^  — a;2  — 2a;  +  2  Gx^  — 5cc  — 6 

17.    — ;r-^ -i 27. 


2:c^-ic  — 1  8x^  — 2a;-15 

(K«-2a;2-a:  +  2  (x - ?/)  (^« - 2/«) 

6x3-23a;-^  +  16a:-3  x^^^f 

6x3-17cc2+lla;  — 2  ^''-a:y  +  2/^ 

a:^  — 2x^  — x2-2a^  +  l  (a«  -  ^*3)  (a^  -  aZ»  +  ^»2) 

To  Reduce  a  Fraction  to  an  Integral  or  Mixed 
Expression. 

Change to  a  mixed  expression. 

(a;3  +  l)-f-  (x  — l)=a;2  +  a;  +  l  +  — ?— •     Hence, 

166.  7/"  the  degree  of  the  numerator  of  a  fraction  equals  or 
exceeds  that  of  the  denominator^  the  fraction  may  be  changed 
to  the  form  of  a  mixed  or  integral  expression  by  dividing  the 
numerator  by  the  denominator. 


FBACTIONS.  107 

If  there  is  a  remainder,  this  remainder  must  be  written 
as  the  numerator  of  a  fraction  of  which  the  divisor  is  the 
denominator,  and  this  fraction  with  its  proper  sign  must 
be  annexed  to  the  integral  part  of  the  quotient. 

Exercise  53. 

Change  to  integral  or  mixed  expressions  : 

^     x^  —  2x  +  l  ^    lOa^-llax-^-lOx" 

1. •  6. • 

X  —  1  oa  —  X 

3a;^  +  2a;  +  l  16(3a:^  +  l) 

X-\-4:  '  4:X  1 

Sx^-]-6x-\-5  2a^  —  5x  —  2 

3.  '—T-j^ 8. 

ic  +  4  x  —  4 

4.  -7-^ 9.   ^• 

a-Yx  a  —  0 

'     (c  — 3  *  bx^-\-4.x  —  l 

To  Reduce  a  Mixed  Expression  to  the  Form  of  a 
Fraction. 

167.   In  arithmetic  5f  means  5  +  f . 

But  in  algebra  the  fraction  connected  with  the  integral 
expression,  as  well  as  the  integral  expression,  may  be  posi- 
tive or  negative  ;  so  that  a  mixed  expression  may  occur  in 
any  one  of  the  following  forms  : 

.a  a  ,  a  a 


108  ALGEBRA. 

Change  ^i  +  -  to  a  fractional  form. 

Since  when  J  is  a  positive  integral  number  there  are 
h  bths  in  1,  in  n  there  will  be  n  times  h  bths,  that  is,  nb  bths, 
which,  with  the  additional  a  bths,  make  (nb  -j-  a)  bths. 


In  like  manner 


and 


,  a      nb-\-a 


a      nb  —  a. 


,  a      — nb-\-a 
a      —  nb  —  a 


—  n 


b  b 

The  same  law  holds  whatever  be  the  value  of  b, 

168.  Hence,  to  reduce  a  mixed  expression  to  a  fraction, 
Multiply  the  integral   expression  by  the  denomi7iator,  to 

the  product  annex  the  numerator,  and  under  the  result  write 
the  denominator. 

169.  It  will  be  seen  that  the  sign  before  the  fraction  is 
transferred  to  the  numerator  when  the  mixed  expression  is 
reduced  to  the  fractional  form,  for  the  denominator  shows 
only  what  part  of  the  numerator  is  to  be  added  or  sub- 
tracted. 

The  dividing  line  has  the  force  of  a  vinculum  or  paren- 
thesis affecting  the  numerator ;  therefore  if  a  minus  sign 
precede  the  dividing  line,  and  this  line  be  removed,  the 
sign  of  every  term,  of  the  numerator  must  be  changed. 
Thus, 

a  —  b C7i  —  (a  —  b) en  —  a-\-b 


(1)    Change  to  fractional  form  x  —  l-\-- 
X  — 1-\ > 

X 

x^  —  x-\-  (x  —  1) 


FRACTIONS.  109 

-1 


X 

x^  —  x-\-x  —  1 


X 

x'-l 


onai  i< 

3rm  X  —  jl  — 

X 

X  — 

-1    ^-\ 

X 

a^- 

-x-(x-l)^ 

X 

x'- 

-x-x-\-l 

X 

-tl 

-2x+l 

Exercise  64. 
Change  to  fractional  form  : 

1     l__?JZi^.  5.    6a-2J 

x-\-y 


Sa^-W 


2.    1+^  6.    a  +  b 


5a  —  6b 
a'  +  b' 


l  +  2a;2  ^     _         2-3a  +  4< 


3.    Sx ■ 7.    7a 


X 


5  —  6a 


a*  +  x«  ^     o         5ax  —  3 

a  —  X  ^d 


110  ALGEBKA. 


9.   — '-^+1.  15.   2a  — b r— • 

a  —  0  a-j-b 

10.   ^-1.  16.  30,-10+     ^^ 


a-\-b  x-{-4: 

"•^-(-  +  y)-  17.   x^  +  .  +  l+^. 

4  ic  — 2 

13.   «-lH ~  19.   a2_2aaj  +  4a;2 ^^ 


a-\-l  a-\-2x 

X  — 15                  ^^                 ,        ,  a^ — ay-Vif 
—'  20.  x  —  a-{-y-\ ^^-LX  . 


Lowest  Common  Denominator. 

170.    To  reduce  fractions  to  equivalent  fractions  having 
the  lowest  common  denominator  : 

Zx    2ii  5 

Reduce  -;-^)  -^^  and  -z—o  to  equivalent  fractions  having 
4:a^    3  a  6a^ 

the  lowest  common  denominator. 

The  L.  C.  M.  of  Aa^  Sa,  and  6a^==12a\ 

Sx 
If  both  terms  of  -—^  be  multiplied  by  3  a,  the  value  of 
4  a 

the  fraction  will  not   be  altered,  but  the   form  will   be 

Q  /TO*  2  ?/ 

changed  to  r-^r—^ ;  if  both  terms  of  ^  be  multiplied  by  4a^, 
x^  a  o  a 

the  equivalent  fraction  r^-^  is  obtained ;  and,  if  both  terms 

5  10 

of  —-Q  be  multiplied  by  2,  the  equivalent  fraction  :rw~s  ^^ 

obtained. 


Hence, 


FRACTIONS.  Ill 


3x       2y_      _5_ 


are  equal  to  12:^ '    12^ '    12^3'  respectively. 

The  multipliers  3  a,  4a^,  and  2  are  obtained  by  dividing 
12  a^,  the  L.  C.  M.  of  the  denominators,  by  the  respective 
denominators  of  the  given  fractions. 

171.  Therefore,  to  reduce  fractions  to  equivalent  frac- 
tions having  the  lowest  common  denominator, 

Find  the  L.  C.  M.  of  the  denominators. 

Divide  the  L.  C.  M.  by  the  denominator  of  each  fraction. 

Multiply  the  first  numerator  hy  the  first  quotient,  the 
second  by  the  second,  quotient,  and  so  on. 

The  products  will  be  the  numerators  of  the  equivale7it 
fractions. 

The  L.  C.  M.  of  the  given  denominators  will  be  the 
denominator  of  each  of  the  equivalent  fractions. 


Exercise  55., 

Reduce  to  equivalent  fractions  with  the  lowest  common 
denominator : 

3x  — 7     4.x  — ^  1  1 

1. —  >    — 77; —  •  5. 


6  18  (a  —  b)(b  —  c)     (a  —  b)(a  —  c) 

2x  —  4?/     Sx  —  Sy  4:0^  xy 

5a^     '       lOx  ^'    3(a-{-b)'   6(a'-b'y 

4a  — 5c     Sa  —  2c  8a:  +  2     2:z;  — 1      3a:  +  2 

3.    — z )      ^^    o 7.    r-)    -; -y 


5ac  12ah  x  —  2  'Sx  —  6     bx  — 10 

5  6  a  —  bm  ^      c  —  bn 

4.    z >  z y  8.    ?  1? 

1  —  X  1  —  x^  mx  nx 


112 


ALGEBRA. 


Addition  and  Subtraction  of  Fractions. 

172.  To  add  fractions: 

Reduce  the  fractions  to  equivalent  fractions  having  the 
lowest  common  denominator. 

Add  the  numerators  of  the  equivalent  fractions. 
Write  the  result  over  the  lowest  common  denominator. 

173.  To  subtract  one  fraction  from  another: 
Reduce  the  fractions   to  equivalent  fractions  having  the 

lowest  common  denominator. 

Subtract  the  numerator  of  the  subtrahend  from  the  numer- 
ator of  the  minuend. 

Write  the  result  over  the  lowest  common  denominator. 

(1)  Simplify,  ___  +  ___. 

The  lowest  common  denominator  (L.  C.  D.)  =  15. 
The  multipliers  are  3  and  1  respectively. 

12aj  -j"  21  =  1st  numerator, 
3  £C  —    4  =  2d  numerator, 
15iC  -f"  17  =  sum  of  numerators. 

.  4a;  +  7      3a;-4^15a;  +  17 
•  *       5       "^     15     ""       15       ' 

/«x    CI-       1-1?        3a  — 45      2a  —  h-\-c  ,  13a  — 4c 

(2)  Simplify, \- — 

TheL.  C.  D.  =  84. 

The  multipliers  are  12,  28,  and  7  respectively. 

36  a  —  48  6  =  1st  numerator, 

—  56a-\-2Sb  —  28  c  =  2d  numerator, 
91a  — 28c  =  3d  numerator. 

71a  —  20  b  —  56  c  =  sum  of  numerators. 
3a  — 4^>      2a  — ^>-fc      13a  — 4c      71a  — 20^>  — 56g 
'''7  3         "^       12       ~  84 


FRACTIONS.  113 

Note.  Since  the  minus  sign  precedes  the  second  fraction,  the 
signs  of  all  the  terms  of  the  numerator  of  this  fraction  are  changed 
after  being  multiplied  by  28. 

Exercise  56. 
Simplify : 

Sx-2y     5x-7tj     8a:4-2y 
5x      "^      10a:     "^       25 

Ax'-Ti/'     3a;-8y      5-2y 
3x^      "^      6x      "^     12     * 


4.a^  +  5b^  ,  3a-i-2b  ,  7-2a 


ix  +  5     3x  —  l 9_ 

3  5x     '^12x'' 


4:X  —  Si/     3a;  +  7      5x  —  2i/     9a;  +  2y 
7       "^     14  21       "^      42      ' 


^    3a;y  — 4      5i/-\-7      6a;^  — 11 


a'  —  2ac-\-c^      b''-2bc-\-c^ 
a^c"  bh' 


5a'-2      Sa'  —  a 
Sa^  8 


a  —  b      b  —  c      c  —  a      aP -\- bc^ -j- ca^ 
cab  abc 


10     J I 1         2a;~^      y-2^ 

•   2irV     6y2;5;      2x^2 "^  4icV  "T  ^^^^  ' 


114  ALGEBRA. 

Simplify  — r-^  H '-^  • 

x-\-y     x  —  y 

The  multipliers  are  x  —  y  and  x  +  y,  respectively. 

x^  —  2xy-\-    2/^  =  1st  numerator. 
x^-\-2xy-\-    y^  =  2d-  numerator. 
2x^  -{-2y^=^  sum  of  numerators. 

or,    2(^2 +  2/2)  =    "    "         " 

''a  +  y      a;  — 2/         a;2  — 2/^ 


Exercise  57. 


Simplify : 


X  —  6      cc  +  5  2a(a-\-x)      2  a  (a  —  x) 


2.  -K—K.        7.     '^         * 


iC  —  T      x  —  S  (a-i-b)b      (a  —  b)a 


3.    ::r-: \-z 8. 


4. 


l  +  £c  '  1  —  x  .   2£c(x  — 1)      4a:(ic  — 2) 

1  2  l  +  g;  1  — g; 

1  — aj      l-a;2*  ^'    l+^^  +  x^      1  — a^  +  a^^' 


1               X  2ax  —  3by       2ax-\-Sby 

x  —  y      (x  —  yf  '   2xy{x  —  y)      2xy(x  +  y) 

,  .    Q.      T-     2a-\-b  2a-b        6ab 
(1)  Simplify 


a  —  b        a-\-b       a^  —  b^ 


The  L.  C.  D.  =  (a  -  b)  {a  +  b). 

The  multipliers  are  a  +  6,  a  —  6,  and  1,  respectively. 


FBACTIONS.  115 

2  a^ -\- 3  ab -\- P  =  1st  numerator, 
—  2a^-{-3ab  —  ^^  =  2d  numerator, 
—  6ab  =3d  numerator, 

0  =  sum  of  numerators. 

2a-}-b      2a  — b         6ab 


a-\-b        a^  —  b^ 


=  0. 


(3)  Simplify  ^.-:-^^+i+,$^. 

The  L.  C.  D.  =  (x  +  y)(x-y)(:x'  +  f). 

The  multipliers  are,  respectively: 
z^  +  2/2,  (X  -  y)  (x2  +  2/2),  (x  +  y){x-  y)  (x2  +  y2),  (X  +  y)  (a:  -   y) 

xh^  -f-  ?/*  =  1st  numerator, 

—  x*  4"  2  irV  ~  2  xhf  -\-  2  xif  —  y^  =  2d  numerator, 

£C*  — y  =  3d  numerator, 

2  ic^</  —  2  j:?/*  =  4th  numerator, 

4  a;*^  —    spy  —  y*  =  sum  of  numerators. 

4  it  V  —  x^ip  —  ?y^ 
.-.  Sum  of  fractions  =  — =^-t — —. — —- 

a*  —  / 


Exercise  58. 
Simplify  : 

1       ,       1       ,      2a  X  X' 


1. 


a      1  —  a       1  —  a* 


l-\-a      1  —  a       1  —  a^  '    1  —  x      1  —  x      l-\-x 


1  —  X      l-\-x      l-^-x"^  y      x-\-y      x^-\-xy 


X  —  1      X  —  2      X  —  3 


__3_      __4a 5a^ 

X  —  a       (x  —  ay       (x  —  a)' 


116 
,        1            1 

ALGEBEA. 
3 

"   x  —  1      x-\-2 
fi            """^          1 

h  —  c                     c  —  a 

x-a      x-h 

(a-by      . 

X  —  b      X  —  a       (x,  —  a)(x  —  b) 
10     ^  +  y         ^^      [     x^y  —  x^ 

y       x-\-y     y(^—f) 


(b  —  c)  (c  —  a)       (c  —  a)  (a  —  b)       (a  —  b)  (b  —  c) 
a^  —  be  ,  b^  —  ac  ,  c^-\-ab 

12. :-— r—  +  '  ' 


13. 


(a  +  b)(a  +  c)   '    (b-\-a)(b-{-c)    '    (c-\-b)(c-{-a) 

a  x  a^-\-x^ 

a  —  X      a-\-2x       (a  —  x')(a-\-2x') 


,4.     .         ,t.         .-.         ,t  .+  ' 


(a  —  b)(b  —  c)       (a  —  b)  (a  —  c)       (a  —  c)  (b  —  c) 
X  —  2y         2x-{-y  2x 


x(x—y)       y(x  +  y)       x^  —  f 


16. 


a  —  b  a  —  b  (a  —  b)(x-\-  y) 


x(a-\-b)       y(a-\-b)  xy(a  +  b) 


Sx  x  +  2y  Sy 

''    {x-^yf      x'^-f'^{x-yy 


a  —  c  a  —  b 

18. 


{a-^-by—c"       (a  +  c)2  — J2 

19       0^  +  ^  Q^  — ^      I   ab(x  —  y) 

ax-\-by      ax  —  by      o?x^  —  b^y^ 


FRACTIONS.  117 

174.   Since  —  =  a,  and — r  =  ^^ 

0  —  0 

it  is  evident  that  if  the  signs  of  both  numerator  and 
denominator  are  changed,  the  value  of  the  fraction  is  not 
altered. 

—  h —  {a  —  h) —  a-\-h h  —  a 


Again, 


G  —  d      —  (c  —  d)       —  c-\-d      d  —  c 


Hence,  if  the  numerator  or  denominator  be  a  compound 
expression,  or  if  both  be  compound  expressions,  the  sign  of 
every  term  in  the  denominator  may  be  changed,  provided 
the  sign  of  every  term  in  the  numerator  be  changed. 

Since  the  change  of  the  sign  before  the  fraction  is  equiv- 
alent to  the  change  of  the  sign  before  every  term  of  the 
numerator  of  the  fraction,  the  sign  before  every  term  of 
the  denominator  Ttiay  he  changed^  'provided  the  sign  before 
the  fraction  he  changed. 

Since,  also,  the  product  of  -|-  a  multiplied  hj  -\-h  is  ah, 
and  the  product  of  —  a  multiplied  by  —  i  is  ah,  the  signs 
of  two  factors,  or  of  any  even  number  of  factors,  of  the 
denominator  of  a  fraction  may  be  changed  without  altering 
the  value  of  the  fraction. 

By  the  application  of  these  principles,  fractions  may 
often  be  changed  to  a  more  simple  form  for  addition  or 
subtraction. 

(1)    Simplify? -2^  +  1^- 

Change  the  signs  before  the  terras  of  the  denominator  of  the  third 
fraction,  and  change  the  sign  before  the  fraction. 
The  result  is, 

2  3  2a;  — 3 

X      2x  —  l      4:x^  —  i 

in  which  the  several  denominators  are  arranged  in  the  same  order  with 
respect  to  x. 


118  ALGEBRA. 

The  L.  C.  I).  =  x(2x  —  l)(2x-}-l). 

Sx^  —  2    =  1st  numerator, 

—  6x^  —  3  03  =  2d  numerator, 

—  2a^-\-Sx  =  M  numerator. 

—  2    =  sum  of  numerators. 

-2 


'.  Sum  of  the  fractions 


x(2x-l)(2x-}-l) 
(2)    Simplify: 

1 + 1 + 1 

Change  the  sign  of  the  factor  {b  —  a)  in  the  denominator  of  the 
second  fraction,  and  change  the  sign  before  the  fraction. 

Then  change  the  signs  of  the  factors  (c  —  a)  and  (c  —  b)  in  the 
denominator  of  the  third  fraction. 

The  result  is, 

1 I + I 

a  (a  —  b)(a  —  c)       b(a  —  b)(b  —  c)       c(a  —  c)(b  —  c) 

in  which  the  factors  of  the  several  denominators  are  written  in 
alphabetical  order. 

TheL.  C.  J).  =  abc(a  —  b)(a  —  c)(b  —  c). 

be  (b  —  c)  =  b^G  —  b(^  =  1st  numerator, 

—  ac(a  —  c)  =  —  a^c  -\-  ac^  =  2d  numerator, 

ab  (a  —  b)  =  a%  —  ab^  =  3d  numerator. 

a^  —  ah  —  ab^  -\-  ac^  -\-  bh  —  bc^  =  sum  of  numerators. 

=  a\b  —  c)  —  a(b^  —  c^)-{-bc(b  —  c), 
=  [a^  —  a(b  +  c)^bc^[b  —  c-], 
=  [a^  —  ab  —  ac-\^  bG~\  [b  —  c~\, 
=  [(a^  —  ac)  —  (ab  —  hc)']\b  —  c], 
=  [a  (a  —  c)  —b  (a  —  c)]  [b  —  c], 
=  (a  —  b)(a  —  c)(b  —  c). 


FBACTIONS. 


119 


.'.  Sum  of  the  fractions 


(a  —  ^)  (c^  —  c)  (b  —  c) 
abc  (a  —  5)  (a  —  c)(b  —  c) 

abc 


Exercise  59. 


Simplify : 


1. 


x  —  y 


3  +  2a;  ,  3a;  — 2   ,  l&x  —  oi? 


3. 


4. 


2-x 
x^ 


2^x 


-4 


x-\-l      1 
1 


x^-1 

4  

3-3y2^2-22/"^6i/  +  6 

1  2 


(2  — m)(3-m)       {m  —  l)(m-^)    '    (m-l)(m-2) 
1  .  1 


aj'  —  b^  '  i^  —  a^  '   a^-\-b^ 


8. 


a 


a  — 2^>      3  a:  (q^  —  ^) 


x  —  b       b-\-x  b'^  —  x^ 

3H-2a;      2  — 3a'   ,  16x— a;^ 


10. 


2— a; 
3 


2  +  a; 

7 


a;2  — 4 
4  — 20a; 


l-2a;      l+2x       4a;2  — 1 


120 

ALGEBRA. 

1 1 

a  +  b 

h-{-c                      G-\-a 

(b-c){c-a)   ' 

(b  —  a)(a  —  c)   '    (a  —  b)(b  —  G) 

12. 

a^  —  he 

b'-{-ac                    c^-i-ab 
^(b  +  e)(b-a)    '    (c-a)(e-\-b) 

(a-b)(a-e) 

13. 

y-\rz 

zA-x                       x-\-ij 

(x  —  y)(x  —  z) 

1    (y_^)(^_^)    1    (^_^)(^_2,) 

14. 

3 

4                            6 

(a-b)(b-e) 

(b  —  a)(G—  a)       (a  —  c){c  —  b) 

1  K 

1 

1            1 

x{x  —  y){x  —  z)       y{y  —  x){y  —  z)       xyz 


Multiplication  of  Fractions. 

175.  In  Arithmetic  it  is  often  necessary  to  take  equal 
parts  of  fractions  of  units. 

Suppose  it  is  required  to  take  §  of  |  of  a  unit. 
Let  the  line  AB  represent  the  unit  of  length. 

vtl    I    I    I    I    I    I    I    I    I    I    I    I    I    I    l« 
C  D  E  F 

Suppose  AB  divided  into  5  equal  parts,  at  C,  D,  E,  and 
F,  and  each  of  these  parts  to  be  subdivided  into  3  equal 
subdivisions. 

Then  one  of  the  parts,  as  AC,  will  contain  3  of  these 
subdivisions,  and  the  whole  line  AB  will  contain  15  of 
these  subdivisions. 

That  is,  ^  of  ^  of  the  line  will  be  ^^3^  of  the  line ; 

J  of  f  will  be  tV  +  tV  +  tV  +  tV^  or  ^%  of  the  line ;  and 

I  of  ^  will  be  twice  ^,  or  j\  of  the  line. 


FRACTIONS.  121 


G  Oi 

Suppose  it  is  required  to  take  -  of  -  of  the  line  AB, 
where  a,  b,  c,  d  represent  positive  whole  numbers. 

^1    I    I    I    I    I    I    I    I    I    I    I    I    I    I    ly? 
C  D  E  F 

Let  the  line  AB  be  divided  into  b  equal  parts,  and  let  each 
of  these  parts  be  subdivided  into  d  equal  subdivisions. 
Then  the  whole  line  will  contain  bd  of  these  subdivisions, 

and  one  of  these  subdivisions  will  be  7-;  of  the  line. 

od 

If  one  of  the  subdivisions  be  taken  from  each  of  a  parts, 
they  will  together  be  —  of  the  line.     That  is, 

and  3  of  -  will  be  c  times  — ,  or  —  of  the  line. 
do  od        od 

Therefore,  to  find  a  fraction  of  a  fraction. 

Find  the  product  of  the  numerators  for  the  numerator  of 
the  product^  and  of  the  denominators  for  the  denominator 
of  the  product. 

176.   NoAV,  -,  X  7  means  -  of  7- 
do  do 

Therefore,  to  find  the  product  of  two  fractions, 

Find  the  product  of  the  numerators  for  the  numerator  of 
the  product,  and  of  the  denominators  for  the  denominator 
of  the  product. 

The  same  rule  will  hold  whatever  be  the  values  of 
a,  b,  c,  d. 

If  a  factor  exist  in  both  a  numerator  and  a  denominator 
it  may  be  cancelled ;  for  the  cancelling  of  a  common  factor 
before  the  multiplication  is  evidently  equivalent  to  cancel- 
ling it  after  the  multiplication ;  and  this  may  be  done  by 
§163. 


122  ALGEBRA. 

Division  of  Fractions. 

177.  Multiplying  by  the  reciprocal  of  a  number  is  equiv- 
alent to  dividing  by  the  number.  Thus,  multiplying  by  J 
is  equivalent  to  dividing  by  4. 

The  reciprocal  of  a  fraction  is  the  fraction  with  its  terms 
interchanged. 

Thus,  the  reciprocal  of  §  is  |,  for  |  X  f  =1.  §  42. 

Therefore,  to  divide  by  a  fraction, 

Interchange  the  terms  of  the  fraction^  and  multiply  by  the 
resulting  fraction.     Thus, 

2a       _1^_3^       2a  _2a 
^^^    S^^^3i~  1   ^  Sx^~  x' 

The  common  factor  cancelled  is  3x. 

^^^l^_^      Ux^_2x 
^^^   27y'^9y~7x^  27f~Sy' 

The  common  factors  cancelled  are  9y  and  Ix. 

ax  ab     (a  -\-x)  (a  —  x)  ax 

^  ^    (a  —  xy  '  a?  —  X?  ab  (a  —  x){a  —  x) 

x(a-\-x^ 

b  (a  —  x) 

The  common  factors  cancelled  are  a  and  a  —  x. 

If  the  divisor  be  an  integral  expression,  it  may  be 
changed  to  the  fractional  form.  §  162. 


Exercise  60. 
Simplify  : 

bx^  d  2p-2  '  p-1 

2x      Sab      Sac  Sx*y  _^  2x? 

^'     a^    c    ^  2b'  '    15ab^  '  Sab^' 


FRACTIONS.  123 

_9^ry^  20a%  25  k^m^      TOn^g      3pm 

10  a'b'c^       IS  xifz  '    \^n\f^lhfrri^Wn 

7.    3^x^-^X-^?^'-         10.     ^-^    x^'~^' 
4x«^      6x2/  2icy^  *    a?-\rah      a^  —  ab 

aM:P_^a--b  x'-\-x-2      a;^-13a;  +  42 

•    a'-b^  '   a-{-b  x'  —  7x  x^-\-2x 

ay^-llx-j-30      x'-Sx  ^^^^      (a±x^ 

x^-6x-\-d        x^-5x  a^-{-x^      (a  —  xf 

^g     2a(x'-y'y^^  ^ 

ex  (x  —  y){x-\'yy 


16. 


a?  +  2ab  ^^ab-2b\  ^^    ^^xy  ^^(x-yY 

a^4-4i^       a^  —  46^  *     x  —  y        x^  —  y^ 


01?  —  4       ic^  —  25  .      m^  —  71?  ^   n  —  m 

x^-\-bx      x'-\-2x  c^-\-d^  ^  c  +  ^' 

g^— 4^  +  3       a^~96t  +  20       g'^  — Ta 
a2_5^_^4X^2_iQ^_^21  ^g*-5a' 

Z>^-7Z>  +  6      ^>^+10Z>  +  24^  b^-\-U 
6^  +  36-4      ^»^-14i  +  48   •   ^^^-S^*^* 

22      «^~?/     X  ^y~'^'''^  X  ^'^^y  - 

x^  —  3xy-\-2f       x^-\-xy        (x  —  yf 

a''—Sa'b-\-Sab^-b^   .   2a5-2^>\^  g'^  +  g^ 
a^  —  b^  3  a  —  b 


124  ALGEBRA. 


(^■■^)2_52  2_(^_^)2 

ic^  +  2ic?/  +  2/^~^^      £c  — 2/  +  « 


Complex  Fractions. 

178.  A  complex  fraction  is  one  whicli  has  a  fraction  in 
the  numerator  or  in  the  denominator,  or  in  both. 

To  simplify  a  complex  fraction, 

Divide  the  numerator  by  the  denominator, 

(1)   Simplify  |- 


|  =  *-i  =  fXi=i. 


(2)   Simplify 

21. 

5i 

2t_ 

5i 

=*=. 

-V-  = 

=  ^\ 

Xf 

=  T¥r. 

(3)    Simplify 

3a; 

Sx 

3a: 

3a;    . 
1     • 

4a;- 
4 

-1_ 

4 
4a;  — 1 

v^^. 

x-i      < 

ta;  — 1 

'^  1 

4 

12  a; 

4a;  — 1 


FRACTIONS.  125 

179.  It  is  often  shorter  to  multiply  both  terms  of  the 
fraction  by  the  L.  C.  D.  of  the  fractions  contained  in  the 
numerator  and  denominator. 

Thus,  in  §  178,  (1),  multiply  both  terms  by  6  ;  (2),  both 
terms  by  24 ;   (3),  both  terms  by  4.     The  results  obtained 

12  X 
are  f,  ^,  4,x  —  V  ^^'^P®^*^^®^^- 

Simplify  by  first  simplifying  the  denominator : 


l  +  a:4- 


1  +  ^4 


1  — x  +  a^ 


X 

■  x(l  —  x-\-x^ 

(l  +  a:)(l-a^  +  a^  + 

X 

X  —  x^-\-a? 


1 


l-{-x-\-a? 
x{l-{-x-\-3(?) 


l-\-X-\-X^—(x  —  X^-^3(^ 

x-\-x^-]rx* 
l^x^ 


The  expression is  reduced  to  the  form 

X  (11  ~~~  X  "^  X^  X  "^  x^  ~\~  x^ 

-TT-. — 77; i — ^. — ,  and  this  is  equal  to  -q— i i — i' 

(l-\-  x)  (1  —  x  -\-  x^  -\-  X  ^  1  +  X  +  iC^ 

X 

The  expression  ^  ,     3  is  reduced  to  the  form 

.         X  "~"  X   "J"  X 

l-\-x-\-x^ 

x(l-\-x-\-x^)  J  xt,-    •  1  i.     a^  +  aj^  +  ic* 

14.^+V-(.-^+^>  and  this  ,s  equal  to  -J^p^- 


126  ALGEBRA. 

Exercise  61. 


Simplify : 

3x  .  X  —  1 


.8.1  ^ 


f(.  +  l)-^-2i  *  1  +  i 


x-l-\ ~  1/  9.    1-f 


2.    5—  l  +  ic-l- 


£C  — 2 


X  —  6 


1-a; 

1 
(/lO.   - 


2a;~2/  1 


X  +  1  2    1^         1  1  +  - 


2      2 


£c  —  a 


11. 


fa      x\ 
\x      a) 


. X  —  a 

x-{-a 


x  —  y      x^  —  f  -J — -A 


x?-\-'if'  I     2x    (xy  —  x^     ^-\-y 


—  x^     x-\-y'\ 
—yf      x—y] 


xy  +  f     x^  +  xy  ^      ^ 

»-|-l^--l  (x^-y')(2x'-2xy) 

^^    a;-l  +  x  +  l  ^^             Hx-yf 

x-\-l x  —  1  '                   xy 

x  —  1      05  +  1  x  +  y 


FBACTI0N8.  127 

obb  ac 


x^ -\- (a -\- b)  X -\- ab      x' -\- (a -\- c)  x -\- ac 

15.     r 

b  —  c 


x'  +  (b-^c)x-\-bc 


a  +  J  b 


X       ,    .  1  ,,*         ^  a-{-b 

''■  r^+  ^^1  1  +  1 


X 


a      b 


2m  —  3-\ —  -tH hr" 

m  ab      ac      be 

18.    _- —  19.    -5 TT-r— N2         20. 

2m  — 1  a^—(b-{-cy 


ah  14      ^ 


1  — a; 

Exercise  62. 

MISCELLANEOUS    EXAMPLES. 

1-    Simplify  ^.^^^s_9^_7^  +  8- 

2.  Find  the  value  of  -^ — ^^ g  .  when  a  =  4,  6  =  i, 

.  a  —  b  —  c  ~\~  Z  oc 

c  =  l. 

2  ab^      c^ 

3.  Find  the  value  of  3a^H 75  when  a  =  4,  Z»  =  4^, 

c  =  l. 

2  11 

4.  Simplify 


(a;2  — 1)2      2a;^  — 4a;  +  2      1  — a^ 

5.  Simplify  (^^  +  i_-^)^(-^^-.--i^). 

X  X 

«     -n..    J  ^1.        1        j^f^  —  ^\     x  —  2a-\-b     ,  a  +  ^ 

6.  Find  the  value  of  I  ; ; ^  when  x  =  —^. 

\x  —  bj      x-\-a  —  2b  2 

»     a-      r^      f    a-\-f>  (^-h      ,      2b'    \ 


a  —  5 

2r* 


128 


ALGEBRA. 


..si.p«,(!;±^-s=|).(j±|-:-=l) 

9.    Simplify 


6-0fe-0-(p-0(: 


)■ 


10.    Simplify 


11.    Simplify 


a-\-x 


a  —  X 
a-\-x 


a^-\-x^ 
a'-x''' 
a^-{-x^ 


12.   Divide  a:«  +  i-3(^|-x^)+4(a.  +  ^)by^  +  ^. 
.  2xy 


13.    Simplify 


2a?2/ 


(^-2/y 


1- 


1  + 


14.   Find  the  value  of  ttt h 


4:ab 


ah 

a  +  b 


2b 


2b-\-x      4:b'  —  x' 


when  X 


X  *4~  ?/  ~~"  1  a  ~T~  1 

15.  Find    the   value   of   ,  ^    when   x  =  — 7  and 

_ab-\-a  ^  '  ' 

^~ab-\-i 

16.  Simplify 

1.1.1 


a  (a  —  b)(a  —  c)       b  (b  —  c)  (b  —  a)       c(c  —  a)  (c  —  b) 
Sabc 


a-1      b—1      c—1 


17.    Simplify 


be-{-ca  —  ab 


a      b       c 


FRACTIONS.  129 


18.    Simplify  — ^ X 


1__1^  m«4-7i» 

7i      m 


U  ' 


19.  Simplify  ^—^11+     ^^^^        j 

20.  Simplify  3a-[*+|2a-(6-c)n  +  |  +  |^ 


21.  Simplify  r :^ ^ ^ — ^-^ ^ 

{a  —  y){a  —  xy      (a  —  x)(a  —  yy 

22.  Simplify ^ 23.    -^i ^-^^-^^ ^ 

24.  Simplify  (^-^--,^J^^_^  +  -,-^^j 

25.  Simplify  ; ^ -  +  - ^^ -  +  ,     ""t/     X 

26.  Simplify  -r — --j 


a  (a  —  b)(a  —  c)       i  (^  —  a)  (b  —  c)       abc 
6  ,       a;4-5 


27.   Simplify  ^iA  X  * 


^ 6_      '■(x-l)(x-2) 

X  — 1 


CHAPTER  IX. 

Pbactional  Equations. 

to  reduce  equations  containing  fractions. 

180.   (1)  1  +  5  =  12. 

Multiply  both  sides  by  4,  the  L.  C.  M.  of  the  denominators. 
Then,  2x  +  x  =  48, 

3a;  =  48. 
.'.z=lQ. 

(2)   ^-4  =  24-1 

Multiply  both  sides  by  24,  the  L.  C.  M.  of  the  denominators. 
Then,  4a;  — 96  =  576  — Sx, 

4a;  +  3a;  =576 +  96, 
7  a;  =  672. 
.-.  X  =  96. 


(3) 

X        X 

3 

11 

L 

-  =x 

-9. 

Multiply  by 

33, 

theL. 

C.  M.  ( 

of  the  denominators. 

Then, 

11a;- 

3x+3 

=  33x-297, 

11 

X  — 3  a;  — 33  a; 

=  -297-3, 

-25x 

=  -300. 

.♦.  X 

=  12. 

Since  the  minus  sign  precedes  the  second  fraction,  in  removing 
the  denominator,  the  +  (understood)  before  x,  the  first  term  of  the 
numerator,  is  changed  to  — ,  and  the  —  before  1,  the  second  term 
of  the  numerator,  is  changed  to  +. 

181.   Therefore,  to  clear  an  equation  of  fractions, 
Multiply  each  term  by  the  L.  C.  M.  of  the  denominators. 


FBACTIONAL   EQUATIONS.  131 

If  a  fraction  is  preceded  by  a  minus  sign,  the  sign  of 
every  term  of  the  numerator  must  be  changed  when  the 
denominator  is  removed. 


Exercise  63. 
Solve  the  equations  : 

1.   50.-^  =  71.  4.   ^-_  =  .--^. 


5  —  2x  ,  ^  6x-S  x  +  2      14      S  +  dx 

5x  +  3      3  — 4a;      a;  _  31      9  —  5x 
'^'        8  3"^2~2  6' 

o  ^ 

7a;  +  5      5a;  — 6      8  — 5a; 
^«-   -6 ^  =  -12- 

a;  +  4      a;  — 4_^  ,  3a;  — 1 

to  o 

13.  i(3a;-4)  +  |(5a;  +  3)  =  43-5a;. 
7  tJ 

14.  i(27-2a;)  =  |-^(7a;-54). 


132  ALGEBRA. 

15.  5x—\Sx  —  S[16  —  6x  —  (4:-5x)-]\=6. 

Bx  —  S      9  —  x      5x   ,  19, 

16.  -^ 3-=T  +  T("-^> 

2x-\-7      9x-  — 8_a;  — 11 

7  11     "      2     * 
8x  —  15      11a;  — l_7a;  +  2 

3  7        ~      13     * 

7a;  +  9      3a;  +  l_9a;  — 13      249  — 9a; 

8  7       ""       4  14       * 


17. 


18. 


19. 


182.  If  the  denominators  contain  both  simple  and  com- 
pound expressions,  it  is  best  to  remove  the  simple  expres- 
sions first,  and  then  each  compound  expression  in  turn. 
After  each  multiplication  the  result  should  be  reduced  to 
the  simplest  form. 

So; 4-5      7a;  — 3  _ 4a;  +  6 

^^^        14     '^6a;  +  2~       7 

Multiply  both  sides  by  14. 

4Q7. 01 

Then,  Sx+6+    ^     ,  ,    -8a;  +  12. 

3a;  +  1 

493; 21 

Transpose  and  combine,      — — -——  =  7. 

oX  "T"  1 

Multiply  by  3x  +  1,  49x  -  21  =  21  x  +  7, 


I8x  =  28. 
•.  X  =  1. 


(2) 


4a;  7a; 

4      ~~4  10     * 


Simplify  the  complex  fractions  by  multiplying  both  terms  of  each 
fraction  by  9. 

27-4X      1      7X-27 
Then,  __  =  -___. 

Multiply  both  sides  by  180. 

135-20x=45-14x-f-54, 
-6x=-36. 
.-.  X  =  6. 


FRACTIONAL   EQUATIONS.  133 


Exercise  64. 

Solve  the  equations: 

9a;  +  20_4(a;  — 3)      x 
^'        36      ~  5ic  — 4  "^4* 

9(2a;-3)       11  a:-l  ^9a;  +  ll 
14        "^  3ic  +  l  ""        7 


3. 


7. 


10. 


10a;  +  17        12a;  +  2  _5a;  — 4 
18  13X-16""       9 

6a;  +  13       3a;  +  5  _2x 

15  5^^-25""  5  ' 

18a:-22   ,  ^      ,   l  +  16a;      ,          101-64a; 
-3^-^  +  2. +  ^^=4^ 24 

^  —  6x        l  —  2x^   ^l  +  3a;      10a;  — 11        1 

16  14(x  — 1)~     21  30       "^105* 

9a;  +  5      8a;-7  ^36a;  +  15      41 
14     "^6ic  +  2~       56       "^56' 

6a!  +  7      2a;  — 2      2a;-}-l 


15 

7a;-6           5 

6a;  +  l 

2a;  — 4       2xi 

15 

Ix-IQ           I 

7a;-6 

X  —  5         X 

35 

6a;-101      5 

183.  Literal  equations  are  equations  in  which  some  or 
all  of  the  known  numbers  are  represented  by  letters ;  the 
numbers  regarded  as  known  numbers  are  usually  repre- 
sented by  the  first  letters  of  the  alphabet. 


134  ALGEBRA. 

(1)  (a  —  x)(a-\-x)=2a^-]-2ax  —  x\ 

Then,  a^  —  x^  =  2a^  +  2ax  —  x% 
—  2 ax-  a2. 
a 

(2)  {x  —  a)  (x  —  h)  —  {x  —  b)  (x  —  g)=^2  (x  —  a) {a  —  c). 

{x^  —  ax  —  bx  +  ah)  —  (x^  —  6x  —  ex  +  6c)  =  2  (ax  —  ex  —  a-  +  ac)^ 

x^  —  ax  —  bx  +  ab  —  x^  +  hx-{-cx  —  bc  =  2ax  —  2cx  —  2a^-{-2ac. 

That  is,  —  3  ax  +  3  ex  =  —  2  a2  +  2  ac  —  a&  +  6c, 

—  3  (a  —  e)  X  =  —  2  a  (a  —  c)  —  6  (a  —  e), 

—  3x=— 2a  —  6. 

2a  +  6 
.'.X= — :: — " 


Exercise  65, 

Solve  the  equations : 

1.  ax-{-hG  =  bx-{-ac.  2.    2  a  —  cx=^3  c  —  5hx. 

3.  a^x -{-hx  —  c  =  Px -\-cx  —  d. 

4.  —  ac^  +  h^c  -\-  abcx  =  ahc  -\-  cmx  —  ac^x  +  hh  —  mc» 

5.  (a -{■  X -\- h)  (a -{- h  —  x)  =  (a -\- x)  (h  —  x)  —  ah. 

6.  {a''-{-xf  =  x^-\-4ta'^a\ 

7.  {a^—x){a^-]-x)  =  a^-\-2ax  —  x\ 


x. 


8. 

ax  —  h.           x-^ac 

\-a  — 

c                        c 

3  a  —  Ja;      1 
.   10.    a^            2        ~2' 

9. 

a(b^x-\-x^)               ,  ax? 

11.    6a              3         - 

T«       ^'-^ 

a  —  X      2x      a 

1^.         ■; 

b                  h           X 

13.?       «V 

x^      4zX  —  ac 

ex 

ax   ,    X       - 

14.    am  —  h -f  —  =  0. 

b       m 


15. 


16. 


FRACTIONAL   EQUATIONS.  135 

Sax  —  2  b      ax  —  a ax      2 

'      3^^  2h~~T~S 

ab-\-x      h"^  —  X      X  —  b       ab  —  x 


b^             a%  a"             ^2 

bx-\-\       a(x^  —  V)  ab      ,     ,    ,  .  1 

17.  ax ^—-=-^ ^  19.    —  =  Jc  +  <Z4.-- 

X                         X  X                                X 

aoi?      .        .   ax       .  ^^     a((P-\-x^            .  ax 

18. [•a-\ =  0.  20.       ^      ' ^=ac-] — -• 

b  —  ex               G  dx                     a 


Exercise  66. 
Solve  the  equations : 

x  —  3    _    x  —  5      .1 

'   4.(x  —  l)~6(x-l)'^9 

2.  _^^(.-2)(.+4), 

x  —  1  x-\-l 

7     ^6x4-1      3(l  +  2a^ 
•   x  —  l~  x  +  1  x'  —  l 

1 1        ^         x-1 

'   2(x  —  3)      3(x  — 2)      (x-'2)(x  —  3) 

2(2a;  +  3)^     6  5x-{-l 

9(7  — a-)   ~7  —  x      4(7  — «)* 

^^_         5(21  +  2.)_ 
0^  +  3  3a:  +  9 

x  —  7      2x  —  W  1 

7. 


£c  +  7       2x~6       2(x-{'7) 
3x-^5^^      2x  +  3 


136  ALGEBRA. 


1320^  +  1   ,  Sx-{-5      ^^  3cc  — 1      4:X  —  2 

9.    -t; rV-H -^=52.      11. 


Sx  +  l     '    x  —  1  2ic  — 1      3^  —  2      6 

10. --] 7^  =  7^ r^-      12.  '       — 


2x~3  '  x  —  2      3cc  +  2  a:  — 1      x  —  1      l  —  x" 

X  —  4      X  —  5      X  —  7      X  —  8 
13. 


X — 5      X — 6      X — 8      x—9 

14.  (x  —  a)(x  —  b)  =  (x--a  —  by. 

15.  (a  — ^)(.'r  — c)  — (^  — c)(a7  — fl'.)  — (c  — a)(x  — ^>)=0. 

16.    :; j-q —  =  i^a;. 

x  — 1  x-\-l 

17.  ^4       ^  ^^ 


x-{-2   '  ic  +  3      ir2_|_5^_pg 
18.    (x-\-iy  =  xl6  —  (l—x)2  —  2. 

25-ix      16^  +  4i_    23 
^  •      0^  +  1    "^    3a:  +  2    ~a;  +  l"^^* 

3a^»c   ,       a^b^       ,    (2a  +  5)^^2^      ^        ,  bx 

20.  — rr  +  T — rTT-s+      /    ■  7X2  =Scx-] 


21. 


4.3  29  2 


ic  — 8   '  2x  — 16      24      3ic— 24 
22. 

x  +  4 


■■■  =-a-o=i-^^^=i=^' 


«,. ' 


24. 


5      x-1 


Uj  If 


J_ 


1(0,-1)   ■  1(0,+  !)  ■ 

15 


0-^) 


CHAPTEE    X. 

Problems. 

Exercise  67. 

Ex.  Find  the  number  tlie  sum  of  whose  third  and  fourth 
parts  is  equal  to  12. 

Let  X  =  the  number. 

Then  -  =  the  third  part  of  the  number, 

and  -  =  the  fourth  part  of  the  number. 

.*.  o  +  7  =  the  sum  of  the  two  parts. 

But  12  =  the  sum  of  the  two  parts. 

.-.^  +  -=12 

Multiply  both  sides  by  12  : 

ix  +  Sx=  144, 
7  X  =  144. 
.-.  X  =  20f 

1.  Find  the  number  whose  third  and  fourth  parts  together 

make  14. 

2.  Find  the  number  whose  third  part  exceeds  its  fourth 

part  by  14. 

3.  The   half,  fourtli,  and  fifth  of  a  certain  number  are 

together  equal  to  76;  find  the  number. 

4.  Find  the  number  whose  double  exceeds  its  half  by  12. 

5.  Divide  60  into  two  such  parts  that  a  seventh  of  one 

part  may  be  equal  to  an  eighth  of  the  other. 


138  ALGEBRA. 

6.  Divide  50  into  two  such  parts  that  a  fourth  of  one 

part  increased  by  five-sixths  of  the  other  part  may 
be  equal  to  40. 

7.  Divide  100  into  two  such  parts  that  a  fourth  of  one 

part  diminished  by  a  third  of  the  other  part  may  be 
equal  to  11. 

8.  The  sum  of  the  fourth,  fifth,  and  sixth  parts  of  a 

certain  number  exceeds  the  half  of  the  number  by 
112.     What  is  the  number? 

9.  The  sum  of  two  numbers  is  5760,  and  their  difference 

is  equal  to  one-third  of  the  greater.  What  are  the 
numbers  ? 

10.  Divide  45  into   two   such  parts   that   the   first  part 

divided  by  2  shall  be  equal  to  the  second  part 
multiplied  by  2. 

11.  Find  a  number  such  that  the  sum  of  its  fifth  and  its 

seventh  parts  shall  exceed  the  difference  of  its  fourth 
and  its  seventh  parts  by  99. 

12.  In  a  mixture  of  wine   and  water,  the  wine  was  25 

gallons  more  than  half  of  the  mixture,  and  the 
water  5  gallons  less  than  one-third  of  the  mixture. 
How  many  gallons  were  there  of  each? 

13.  In  a  certain  weight  of  gunpowder  the  saltpetre  was  6 

pounds  more  than  half  of  the  weight,  the  sulphur 
5  pounds  less  than  the  third,  and  the  charcoal  3 
pounds  less  than  the  fourth  of  the  weight.  How 
many  pounds  were  there  of  each? 

14.  Divide  46   into  two  parts  such  that  if  one  part  be 

divided  by  7,  and  the  other  by  3,  the  sum  of  the 
quotients  shall  be  10. 


PROBLEMS.  139 

16.  A  house  and  garden  cost  $850,  and  five  times  the  price 
of  the  house  was  equal  to  twelve  times  the  price  of 
the  garden.     What  was  the  price  of  each  ? 

16.  A  man  left  the  half   of  his  property  to  his  wife,  a 

sixth  to  each  of  his  two  children,  a  twelfth  to  his 
brother,  and  the  remainder,  amounting  to  $600,  to 
his  sister.     What  was  the  amount  of  his  property  ? 

17.  The  sum  of  two  numbers  is  a  and  their  difference  is  b] 

find  the  numbers. 

18.  Find  two  numbers  of  which  the  sum  is  70,  such  that 

the  first  divided  by  the  second  gives  2  as  a  quotient 
and  1  as  a  remainder. 

19.  Find  two  numbers  of  which  the  difference  is  25,  such 

that  the  second  divided  by  the  first  gives  4  as  a 
quotient  and  4  as  a  remainder. 

20.  Divide  the  number  208  into  two  parts  such  that  the 

sum  of  the  fourth  of  the  greater  and  the  third  of 
the  smaller  is  less  by  4  than  four  times  the  difference 
of  the  two  parts. 

21.  Find  four  consecutive  numbers  whose  sum  is  82. 

Note  I.  It  is  to  be  remembered  that  if  x  represents  a  person's  age 
at  the  present  time,  his  age  a  years  ago  will  be  represented  by  x  —  a, 
and  a  years  hence  by  x  +  a. 

Ex.  In  eight  years  a  boy  will  be  three  times  as  old  as  he 
was  eight  years  ago.     How  old  is  he  ? 

Let  X  =  the  number  of  years  of  his  age. 

Then  x  —  8  =  the  number  of  years  of  his  age  eight  years  ago, 
and        X  +  8  =  the  number  of  years  of  his  age  eight  years  hence. 
.-.  X  +  8  =  3  (x  —  8), 
x  +  8  =  3x— 24, 
X  — 3x=  —  24  — 8, 
-2x=  -32, 
x=16. 


140  ALGEBRA. 

22.  A  is  72  years  old,  and  B's  age  is  two-thirds  of  A's. 

How  long  is  it  since  A  was  five  times  as  old  as  B  ? 

23.  A  mother  is  70  years  old,  her  daughter  is  half  that 

age.    How  long  is  it  since  the  mother  was  three  and 
one-third  times  as  old  as  the  daughter  ? 

24.  A  father  is  three  times  as  old  as  the  son ;  four  years 

ago  the  father  was  four  times  as  old  as  the  son  then 
was.     What  is  the  age  of  each  ? 

25.  A  is  twice  as  old  as  B,  and  seven  years  ago  their 

united   ages   amounted  to  as  many  years  as  now 
represent  the  age  of  A.     Eind  the  ages  of  A  and  B. 

26.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  25  years  ;  the  difference  is  one-third  what 
the  sum  will  be  in  20  years.  What  is  the  age  of  each? 

Note  II.  If  A  can  do  a  piece  of  work  in  x  days,  the  part  of  the 
work  that  he  can  do  in  one  day  will  be  represented  by  i.  Thus,  if  he 
can  do  the  work  in  5  days,  in  1  day  he  can  do  \  of  the  work. 

Ex.  A  can  do  a  piece  of  work  in  5  days,  and  B  can  do  it 
in  4  days.  How  long  will  it  take  A  and  B  together 
to  do  the  work? 

Let  X  =  the  number  of  days  it  will  take  A  and  B  together. 

Then        i  =  the  part  they  can  do  in  one  day. 
Now,        \  =  the  part  A  can  do  in  one  day, 
and  \  =  the  part  B  can  do  in  one  day. 

.*.    i  +  $  =  the  part  A  and  B  can  do  in  one  day  together. 

•••  i  +  i  =  -^ 
^x-¥  5ic  =  20, 

9a;  =  20, 

x  =  2f. 

Therefore  they  will  do  the  work  in  2|  days. 

27.  A  can  do  a  piece  of  work  in  5  days,  B  in  6  days,  and 

C  in  7J-  days ;  in  what  time  will  they  do  it,  all 
working  together? 


PROBLEMS.  141 

28.  A  can  do  a  piece  of  work  in  2^  days,  B  in  3^  days,  and 

C  in  3f  days  J  in  what  time  will  they  do  it,  all 
working  together? 

29.  Two  men  who  can  separately  do  a  piece  of  work  in  15 

days  and  16  days,  can,  with  the  help  of  another,  do 
it  in  6  days.  How  long  would  it  take  the  third  man 
to  do  it  alone? 

30.  A  can  do  half  as  much  work  as  B,  B  can  do  half  as 

much  as  C,  and  together  they  can  complete  a  piece 
of  work  in  24  days.  In  what  time  can  each  alone 
complete  the  work? 

31.  A  does  I  of  a  piece  of  work  in  10  days,  when  B  comes 

to  help  him,  and  they  finish  the  work  in  3  days 
more.  How  long  would  it  have  taken  B  alone  to 
do  the  whole  work? 

32.  A  and  B  together  can  reap  a  field  in  12  hours,  A  and 

C  in  16  hours,  and  A  by  himself  in  20  hours.  In 
what  time  can  B  and  C  together  reap  it  ?  In  what 
time  can  A,  B,  and  C  together  reap  it  ? 

33.  A  and  B  together  can  do  a  piece  of  work  in  12  days, 

A  and  C  in  15  days,  B  and  C  in  20  days.  In  what 
time  can  they  do  it,  all  working  together  ? 

Note  III.  If  a  pipe  can  fill  a  vessel  in  x  hours,  the  part  of  the 
vessel  filled  by  it  in  one  hour  will  be  represented  by  i.  Thus,  if  a 
pipe  will  fill  a  vessel  in  3  hours,  in  1  hour  it  will  fill  \  of  the  vessel. 

34.  A  tank  can  be  filled  by  two  pipes  in  24  minutes  and 

30  minutes,  respectively,  and  emptied  by  a  third  in 
20  minutes.  In  what  time  will  it  be  filled  if  all 
three  are  running  together? 

35.  A  tank  can  be  filled  in  15  minutes  by  two  pipes,  A  and 

B,  running  together.     After  A  has  been  running  by 


142  ALGEBRA. 

itself  for  5  minutes,  B  is  also  turned  on,  and  tlie 
tank  is  filled  in  13  minutes  more.  In  what  time 
may  it  be  filled  hj  each  pipe  separately  ? 

36.  A  cistern  could  be  filled  by  two  pipes  in  6  hours  and  8 

hours,  respectively,  and  could  be  emptied  by  a  third 
in  12  hours.  In  what  time  would  the  cistern  be 
filled  if  the  pipes  were  all  running  together? 

37.  A  tank  can  be  filled  by  three  pipes  in  1  hour  and 

20  minutes,  3  hours  and  20  minutes,  and  5  hours, 
respectively.  In  what  time  will  the  tank  be  filled 
when  all  three  pipes  are  running  together? 

38.  If  three  pipes  can  fill  a  cistern  in  a,  b,  and  c  minutes, 

respectively,  in  what  time  will  it  be  filled  by  all 
three  running  together? 

39.  The  capacity  of  a  cistern  is  755^  gallons.    The  cistern 

has  three  pipes,  of  which  the  first  lets  in  12  gallons 
in  3J  minutes,  the  second  15^  gallons  in  2|  minutes, 
the  third  17  gallons  in  3  minutes.  In  what  time 
will  the  cistern  be  filled  by  the  three  pipes  running 
together  ? 

Note  IV.     In  questions  involving  distance,  time,  and  rate : 

Distance      ^. 
-— — —  =  Time. 
Rate 

Thus,  if  a  man  travels  40  miles  at  the  rate  of  4  miles  an  hour, 

40 

—  =  the  number  of  hours  required. 

Ex.  A  courier  who  goes  at  the  rate  of  31^  miles  in  5 
hours,  is  followed,  after  8  hours,  by  another  who 
goes  at  the  rate  of  22^  miles  in  3  hours.  In  how 
many  hours  will  the  second  overtake  the  first? 

Since  the  first  goes  31^  miles  in  5  hours,  his  rate  per  hour  is  6^ 
miles. 


PROBLEMS.  143 

Since  the  second  goes  22^  miles  in  3  hours,  hi^rate  per  hour  is  7^ 
miles. 

Let  X  =  the  number  of  hours  the  first  is  travelling. 

Then  a;  —  8  =  the  number  of  hours  the  second  is  travelling. 

Then  6^j^x  =  the  number  of  miles  the  first  travels ; 

(x  —  8)  7 1  =  the  number  of  miles  the  second  travels. 
They  both  travel  the  same  distance. 

...  6^a:=(x-8)7i. 

The  solution  of  which  gives  42  hours. 

40.  A  sets  out  and  travels  at  the  rate  of  7  miles  in  6 

hours.  Eight  hours  afterwards,  B  sets  out  from  the 
same  place  and  travels  in  the  same  direction,  at  the 
rate  of  5  miles  in  3  hours.  In  how  many  hours  will 
B  overtake  A? 

41.  A  person  walks  to  the  top  of  a  mountain  at  the  rate 

of  2^  miles  an  hour,  and  down  the  same  way  at  the 
rate  of  3^  miles  an  hour,  and  is  out  5  hours.  How 
far  is  it  to  the  top  of  the  mountain? 

42.  A  person  has  a  hours  at  his  disposal.     How  far  may 

he  ride  in  a  coach  which  travels  b  miles  an  hour,  so 
as  to  return  home  in  time,  walking  back  at  the  rate 
of  c  miles  an  hour  ? 

43.  The  distance  between  London  and  Edinburgh  is  360 

miles.  One  traveller  starts  from  Edinburgh  and 
travels  at  the  rate  of  10  miles  an  hour ;  another 
starts  at  the  same  time  from  London,  and  travels  at 
the  rate  of  8  miles  an  hour.  How  far  from  London 
will  they  meet? 

44.  Two  persons  set  out  from  the  same  place  in  opposite 

directions.  The  rate  of  one  of  them  per  hour  is  a 
mile  less  than  double  that  of  the  other,  and  in  4 
hours  they  are  32  miles  apart.  Determine  their 
rates. 


144 


ALGEBRA. 


45.  In  going  a  certain  distance,  a  train  travelling  35  miles 
an  hour  takes  2  hours  less  than  one  travelling  25 
miles  an  hour.     Determine  the  distance. 

Note  V.     In  problems  relating  to  clocks,  it  is  to  be  observed  that 
the  minute-hand  moves  twelve  times  as  fast  as  the  hour-hand. 

Ex.    Find  the  time  between  two  and  three  o'clock  when 
the  hands  of  a  clock  are  : 
I.    Together. 
II.    At  right  angles  to  each  other. 
III.    Opposite  to  each  other. 


Fig.  1. 


Fig.  2. 


Fig.  3. 


I.  Let  CH  and  CM  (Fig.  1)  denote  the  positions  of  the  hour  and 
minute  hands  at  2  o'clock,  and  CB  the  position  of  both  hands  when 
together. 

Then  arc  HB  =  one-twelfth  of  arc  MB. 

Let  X  =  the  number  of  minute-spaces  in  arc  MB. 


Then 


rr  =  the  number  of  minute-spaces  in  arc  HB, 


and  10  =  the  number  of  minute-spaces  in  arc  MH. 

Now  arc  MB  =  the  arc  MH  +  arc  HB. 

X 


That  is, 


x=  10  + 


12 


The  solution  of  this  equation  gives  x  =  10|-f . 
Hence,  the  time  is  10^^  minutes  past  2  o'clock. 

II.    Let  CB  and  CD  (Fig.  2)  denote  the  positions  of  the  hour  and 
minute  hands  when  at  right  angles  to  each  other. 


PROBLEMS.  146 

Let  X  =  the  number  of  minute-spaces  in  arc  MHBD. 

X 

Then         —  =  the  number  of  minute-spaces  in  arc  HB^ 

and  10  =  the  number  of  minute-spaces  in  arc  MH. 

15  =  the  number  of  minute-spaces  in  arc  BB. 
Now  arc  MHBD  =  the  arcs  MH  +  HB -h  BB. 

That  is,  X  =  10  +  ^4  +  15- 

The  solution  of  this  equation  gives  x  =  27y\. 
Hence,  the  time  is  27y\  minutes  past  2  o'clock. 

III.   Let  CB  and  CB  (Fig.  3)  denote  the  positions  of  the  hour  and 
minute  hands  when  opposite  to  each  other. 

Let  X  =  the  number  of  minute-spaces  in  arc  MHBB. 

X 

Then         —  =  the  number  of  minute-spaces  in  arc  HB, 

and  10  =  the  number  of  minute-spaces  in  arc  MH, 

30  =  the  number  of  minute-spaces  in  arc  BB. 
Now  arc  MHBB  =  the  arcs  MH  -h  HB  +  BB. 

That  is,  X  =  10  +  :^  +  30. 

12 

The  solution  of  this  equation  gives  x  =  43/^. 
Hence,  the  time  is  43y^y  minutes  past  2  o'clock. 

46.  At  what  time  are  the  hands  of  a  watch  together : 

I.    Between  3  and  4  ? 

II.    Between  6  and  7  ? 

III.    Between  9  and  10  ? 

47.  At  what  time  are  the  hands  of  a  watch  at  right-angles  : 

T.    Between  3  and  4  ? 

II.    Between  4  and  5  ? 

III.    Between  7  and  8? 

48.  At  what  time  are  the  hands  of  a  watch  opposite  to 

each  other : 

I.  Between  1  and  2? 

II.  Between  4  and  5? 

III.  Between  8  and  9  ? 


146  ALGEBRA. 

49.  It  is  between  2  and  3  o'clock;  but  a  person  looking 
at  his  watch  and  mistaking  the  hour-hand  for  the 
minute-hand,  fancies  that  the  time  of  day  is  55 
minutes  earlier  than  it  really  is.  What  is  the  true 
time? 

Note  VI.  It  is  to  be  observed  that  if  a  represent  the  number  of 
feet  in  the  length  of  a  step  or  leap,  and  x  the  number  of  steps  or  leaps 
taken,  then  oar  will  represent  the  number  of  feet  in  the  distance 
made. 

Ex.  A  hare  takes  4  leaps  to  a  greyhound's  3  ;  but  2  of  the 
greyhound's  leaps  are  equivalent  to  3  of  the  hare's. 
The  hare  has  a  start  of  50  leaps.  How  many  leaps 
must  the  greyhound  take  to  catch  the  hare  ? 

Let     3  X  =  the  number  of  leaps  taken  by  the  greyhound. 
Then  4x  =  the  number  of  leaps  of  the  hare  in  the  same  time. 
Also,  let  a  denote  the  number  of  feet  in  one  leap  of  the  hare. 

q  ^ 

Then  —  will  denote  the  number  of  feet  in  one  leap  of  the  grey- 
hound. 

That  is,  3  X  X  — -  =  the  whole  distance, 

and  (50  +  4  x)  a  =  the  whole  distance. 

.-.   ^=(50-l-4x)a. 

9x 
Divide  by  a,  —  =  50  -}-  4x, 

A 

9x=  100  +  8a;. 
X  =  100. 
.-.  3x  =  300. 

Thus  the  greyhound  must  take  300  leaps. 

50.  A  hare  takes  6  leaps  to  a  dog's  5,  and  7  of  the  dog's 
leaps  are  equivalent  to  9  of  the  hare's.  The  hare 
has  a  start  of  50  of  her  own  leaps.  How  many 
leaps  will  the  hare  take  before  she  is  caught? 


PROBLEMS.  147 

61.  A  greyhound  makes  3  leaps  while  a  hare  makes  4;  but 
2  of  the  greyhound's  leaps  are  equivalent  to  3  of 
the  hare's.  The  hare  has  a  start  of  50  of  the  grey- 
hound's leaps.  How  many  leaps  does  each  take 
before  the  hare  is  caught? 

52.  A  greyhound  makes  2  leaps  while  a  hare  makes  3;  but 

1  leap  of  the  greyhound  is  equivalent  to  2  of  the 
hare's.  The  hare  has  a  start  of  80  of  her  own  leaps. 
How  many  leaps  will  the  hare  take  before  she  is 
caught  ? 

Note  VII.  It  is  to  be  observed  that  if  the  number  of  units  in  the 
breadth  and  length  of  a  rectangle  be  represented  by  x  and  x  +  a, 
respectively,  then  x  {x  +  a)  will  represent  the  number  of  surface  units 
in  the  rectangle,  the  unit  of  surface  having  the  same  name  as  the 
linear  unit  in  which  the  sides  of  the  rectangle  are  expressed. 

53.  A  rectangle  whose   length  is  5  feet   more   than   its 

breadth  would  have  its  area  increased  by  22  feet  if 
its  length  and  breadth  were  each  made  a  foot  more. 
Find  its  dimensions. 

54.  A  rectangle  has  its  length  and  breadth  respectively  5 

feet  longer  and  3  feet  shorter  than  the  side  of  the 
equivalent  square.     Find  its  area. 

55.  The  length  of  a  rectangle  is  an  inch  less  than  double 

its  breadth ;  and  when  a  strip  3  inches  wide  is  cut 
off  all  round,  the  area  is  diminished  by  210  inches. 
Find  the  size  of  the  rectangle  at  first. 

56.  The  length  of  a  floor  exceeds  the  breadth  by  4  feet ;  if 

each  dimension  were  increased  by  1  foot,  the  area 
of  the  room  v/ould  be  increased  by  27  square  feet. 
Find  its  dimensions. 

Note  VIII.  It  is  to  be  observed  that  if  h  pounds  of  metal  lose  a 
pounds  when  weighed  in  water,  1  pound  will  lose  i  of  a  pounds, 

?    nf    a    ■nniTnrl 


148  ALGEBRA. 

57.  A  mass  of  tin  and  lead  weighing  180  pounds  loses  21 

pounds  when  weighed  in  water ;  and  it  is  known  that 
37  pounds  of  tin  lose  5  pounds,  and  23  pounds  of 
lead  lose  2  pounds,  when  weighed  in  water.  How 
many  pounds  of  tin  and  of  lead  in  the  mass  ? 

58.  If  19  pounds  of  gold  lose  1  pound,  and  10  pounds  of 

silver  lose  1  pound,  when  weighed  in  water,  find  the 
amount  of  each  in  a  mass  of  gold  and  silver  weighing 
106  pounds  in  air  and  99  pounds  in  water. 

59.  Fifteen  sovereigns  should  weigh  77  pennyweights  ;  but 

a  parcel  of  light  sovereigns,  having  been  weighed  and 
counted,  was  found  to  contain  9  more  than  was  sup- 
posed from  the  weight ;  and  it  appeared  that  21  of 
these  coins  weighed  the  same  as  20  true  sovereigns. 
How  many  were  there  all  together  ? 


60.  There  are  two  silver  cups,  and  one  cover  for  both.    The 

first  weighs  12  ounces,  and  with  the  cover  weighs 
twice  as  much  as  the  other  without  it;  but  the  sec- 
ond with  the  cover  weighs  one-third  more  than  the 
first  without  it.     Find  the  weight  of  the  cover. 

61.  A  man  wishes  to  enclose  a  circular  piece  of  ground  with 

palisades,  and  finds  that  if  he  sets  them  a  foot  apart 
he  will  have  too  few  by  150  ;  but  if  he  sets  them  a 
yard  apart  he  will  have  too  many  by  70.  What  is  the 
circuit  of  the  piece  of  ground  ? 

62.  A  horse  was  sold  at  a  loss  for  $200  ;  but  if  it  had  been 

sold  for  $250,  the  gain  would  have  been  three-fourths 
of  the  loss  when  sold  for  $200.  Find  the  value  of 
the  horse. 

63.  A  and  B  shoot  by  turns  at  a  target.     A  puts  7  bullets 

out  of.  12,  and  B  9  out  of  12,  into  the  centre.  Be- 
tween them  they  put  in  32  bullets.  How  many  shots 
did  each  fire  ? 


PROBLEMS.  149 

64.  A  boy  buys  a  number  of  apples  at  the  rate  of  5  for  2 

pence.  He  sells  half  of  them  at  2  a  penny  and  the 
rest  at  3  a  penny,  and  clears  a  penny  by  the  trans- 
action.    How  many  does  he  buy  ? 

65.  A  person  bought  a  piece  of  land  for  $6750,  of  which 

he  kept  |  for  himself.  At  the  cost  of  $250  he  made 
a  road  which  took  j\  of  the  remainder,  and  then  sold 
the  rest  at  12^  cents  a  square  yard  more  than  double 
the  price  it  cost  him,  thus  clearing  his  outlay  and 
$500  besides.  How  much  land  did  he  buy,  and 
what  was  the  cost  price  per  yard? 

66.  A  boy  who  runs  at  the  rate  of  12  yards  per  second 

starts  20  yards  behind  another  whose  rate  is  10^ 
yards  per  second.  How  soon  will  the  first  boy  be 
10  yards  ahead  of  the  second  ? 

67.  A  merchant  adds  yearly  to  his  capital  one-third  of  it, 

but  takes  from  it  at  the  end  of  each  year,  $5000  for 
expenses.  At  the  end  of  the  third  year,  after  de- 
ducting the  last  $5000,  he  has  twice  his  original 
capital.     How  much  had  he  at  first  ? 

68.  A  shepherd  lost  a  number  of  sheep  equal  to  one-fourth 

of  his  flock  and  one-fourth  of  a  sheep  j  then,  he  lost 
a  number  equal  to  one-third  of  what  he  had  left  and 
one-third  of  a  sheep  ;  finally  he  lost  a  number  equal 
to  one-half  of  what  now  remained  and  one-half  a 
sheep,  after  which  he  had  but  25  sheep  left.  How 
many  had  he  at  first  ? 

69.  A  trader  maintained  himself  for  three  years  at  an  ex- 

pense of  $250  a  year  ;  and  each  year  increased  that 
part  of  his  stock  which  was  not  so  expended  by  one- 
third  of  it.  At  the  end  of  the  third  year  his  original 
stock  was  doubled.    What  was  his  original  stock  ? 


150  ALGEBRA. 

70.  A  cask  contains  12  gallons  of  wine   and  18   gallons 

of  water ;  another  cask  contains  9  gallons  of  wine 
and  3  gallons  of  water.  How  many  gallons  must 
be  drawn  from  each  cask  to  produce  a  mixture 
containing  7  gallons  of  wine  and  7  gallons  of  water? 

71.  The  members  of  a  club  subscribe  each  as  many  dollars 

as  there  are  members.  If  there  had  been  12  more 
members,  the  subscription  from  each  would  have 
been  $10  less,  to  amount  to  the  same  sum.  How 
many  members  were  there  ? 

72.  A  number  of  troops  being  formed  into  a  solid  square, 

it  was  found  there  were  60  men  over;  but  when 
formed  in  a  column  with  5  men  more  in  front  than 
before,  and  3  men  less  in  depth,  there  was  lacking 
one  man  to  complete  it.    Find  the  number  of  troops. 

73.  An  officer  can  form  the  men  of  his  regiment  into  a 

hollow  square  twelve  deep.  The  number  of  men  in 
the  regiment  is  1296.  Find  the  number  of  men  in 
front  of  the  hollow  square. 

74.  A  person  starts  from  P  and  walks  towards  Q  at  the 

rate  of  3  miles  an  hour;  20  minutes  later  another 
person  starts  from  Q  and  walks  towards  P  at  the 
rate  of  4  miles  an  hour.  The  distance  from  P  to  Q 
is  20  miles.     How  far  from  P  will  they  meet  ? 

75.  A  person  engaged  to  work  a  days  on  these  conditions  : 

for  each  day  he  worked  he  was  to  receive  h  cents, 
and  for  each  day  he  was  idle  he  Was  to  forfeit  c 
cents.  At  the  end  of  a  days  he  received  d  cents. 
How  many  days  was  he  idle  ? 

76.  A  banker  has  two  kinds  of  coins  :  it  takes  a  pieces  of 

the  first  to  make  a  dollar,  and  b  pieces  of  the  second 
to  make  a  dollar.  A  person  wishes  to  obtain  c 
pieces  for  a  dollar.  How  many  pieces  of  each  kind 
must  the  banker  give  him  ? 


CHAPTEE   XI. 
Simultaneous  Equations  of  the  Eirst  Degree. 

184.  If  one  equation  contain  two  unknown  numbers,  an 
indefinite  number  of  pairs  of  values  may  be  found  that  will 
satisfy  the  equation. 

Thus,  in  the  equation  x-\-  y=^  10,  any  values  may  be 
given  to  sc,  and  corresponding  values  for  y  may  be  found. 
Any  pair  of  these  values  substituted  for  x  and  y  will 
satisfy  the  equation. 

186.  But  if  a  second  equation  be  given,  expressing  dif- 
ferent relations  between  the  unknown  numbers,  only  ons 
pair  of  values  of  x  and  y  can  be  found  that  will  satisfy 
both  equations. 

Thus,  if  besides  the  equation  x-\-y  =  10,  another  equa- 
tion, aj  —  ?/  =  2,  be  given,  it  is  evident  that  the  values  of 
X  and  y  which  will  satisfy  both  equations  are 


for  6  +  4  =  10,   and  6  —  4  =  2;    and  these  are  the  only 
values  of  x  and  y  that  will  satisfy  both  equations. 

186.  Equations  that  express  different  relations  between 
the  unknown  numbers  are  called  independent  equations. 

Thus,  x-\-y  =  10  and  x  —  y^2  are  independent  equa- 
tions; they  express  different  relations  between  x  and  y. 
But  x-\-y  =  l^  and  3a;-f  3?/  =  30  are   not   independent 


152  ALGEBRA. 

equations  ;  one  is  derived  immediately  from  the  other, 
and  both  express  the  same  relation  between  the  unknown 
numbers. 

187.  Equations  that  are  to  be  satisfied  by  the  same 
values  of  the  unknown  numbers  are  called  simultaneous 
equations. 

188.  Simultaneous  equations  are  solved  by  combining 
the  equations  so  as  to  obtain  a  single  equation  containing 
only  one  unknown  number  ;  and  this  process  is  called 
elimination. 

Three  methods  of  elimination  are  generally  given : 
I.    By  Addition  or  Subtraction. 
II.    By  Substitution. 
III.    By  Comparison. 

Elimination  by  Addition  or  Subtraction. 

(1)    Solve:  2x  —  3y=    4|  (1) 

3x  +  22/  =  32j  (2) 

Multiply  (1)  by  2  and  (2)  by  3, 

4ic-6y=      8  (3) 

9a;  +  6y=    96  (4) 

Add  (3)  and  (4),  13  a:  =  104 

.-.  X  =  8. 
Substitute  the  value  of  x  in  (2), 

24  +  2  ?/  =  32. 
.-.  ?/  =  4. 
In  this  solution  y  is  eliminated  by  addition. 

(2). Solve:  6cc  +  35?/r=177\  (1) 

^x  —  21y=    33J  (2) 

Multiply  (1)  by  4  and  (2)  by  3, 

24  a;  +  140  ?/  =  708  (3) 

24  X—    63  y=    99  (4) 

Subtract  (4)  from  (3),  203  y  =  609 

.-.?/  =  3. 


SIMULTANEOUS    EQUATIONS.  153 

Substitute  the  value  of  y  in  (2). 
8x-63  =  33, 
.-.  X  =  12. 
In  this  solution  x  is  eliminated  by  subtraction. 

189.  E^ence,  to  eliminate  an  unknown  number  by  addi- 
tion or  subtraction, 

Multiply  the  equations  hy  such  numbers  as  will  make  the 
coefficients  of  the  symbol  for  this  unknown  number  equal  in 
the  resulting  equations. 

Add  the  resulting  equations,  or  subtract  one  from  the  other, 
according  as  these  equal  coefficients  have  unlike  or  like  signs. 

Note.  It  is  generally  best  to  select  that  unknown  number  to  be 
eliminated  which  requires  the  smallest  multipliers  to  make  its  co- 
efficients equal ;  and  the  smallest  multiplier  for  each  equation  is  found 
by  dividing  the  L.  C.  M.  of  the  coefficients  of  this  unknown  number 
by  the  given  coefficient  in  that  equation.  Thus,  in  example  (2),  the 
L.  C.  M.  of  6  and  8  (the  coefficients  of  x),  is  24,  and  hence  the  smallest 
multipliers  of  the  two  equations  are  4  and  3  respectively. 

Sometimes  the  solution  is  simplified  by  first  adding  the 
given  equations,  or  by  subtracting  one  from  the  other. 

(3) 


X 

+  49y=    51 

(1) 

49  X 

+      y=    99 

(2) 

Add  (1)  and  (2), 

60  X 

+  50y=150 

(8) 

Divide  (3)  by  50, 

x  +  y  =  Z. 

W 

Subtract  (4)  from  (I), 

48y=:48. 
.'.y=\. 

Subtract  (4)  from  (2), 

48x=96. 
.-.  X  =  2. 

Exercise  68. 
Solve  by  addition  or  subtraction : 

.      1.  2ic  +  3?/  =  7|     3.  7a;4-2?/  =  30'l  5.  5x  +  47/  =  58\ 

4cc  — 5?/  =  3J    ^          y  —  2,x=   2j  8x-{-7y  =  67J 

^2.     x  —  2y  =  4:\^      4.  3x  — 5y  =  51"\  6.  3£c  +  2?/  =  39\ 

2x—   y  =  5j           2x  +  7y=   3J  3?/-2a;  =  13j 


154  ALGEBBA. 


7.  Sx  —  4:y  =  —  5^  11.    12a;+    7?/  =  176\ 
4:x  —  5y=     Ij  3?/  — 19a^=     3J 

8.  llcc  +  32/  =  100\  12.    2x  —  l2j=   8\ 

4cc  — 7?/=      4J  4?/  — 9ic  =  19j 

9.  a;  +  492/  =  693\  13.    692/  — 17£c=    103\ 
49a;+      y  =  357j  14aj-13y==-41  J 

10.    17cc  +  3?/  =  57\  14.    17a:  +  30?/  =  59\ 
16?/  — 3£c  =  23J  19ir  +  282/  =  77j 


Elimination  by  Substitution. 


Solve:  2x-\-^y  = 


Zx-\-ly 


:?} 


2x  +  3?/  =  8  (1) 

3  X  +  7  ?/  =  7  (2) 

Transpose  3  2/  in  (1),  2  x  =  8  —  3  y.  (3) 

Divide  by  coefficient  of  x,         x  —  — -—^  (4) 


Substitute  the  value  of  x  in  (2),  3  (  — -—^  j  +  1  y—1^ 

24  — 9?/  ,   ^ 

24  —  9^+14?/=  14, 

5  ?/  =  -  10. 
•••2/=  -2. 
Substitute  the  value  of  2/  in  (1),  2  x  —  6  =  8. 

.-.  x=7. 

190.  Hence  to  eliminate  an  unknown  number  by  sub- 
stitution, 

From  one  of  the  equations  obtain  the  value  of  one  of  the 
unknown  numbers  in  terms  of  the  other. 

Substitute  for  this  unknown  number  its  value  in  the  other 
equation,  and  reduce  the  resulting  equation. 


SIMULTANEOUS    EQUATIONS.  155 


Exercise  69. 

Solve  by  substitution : 

.    1.    3x-4.y  =  2\ 
7x-9y  =  7) 

8. 

3a^-4i/  =  18| 

3x  +  2y=   OJ 

^  2.    7x-57/  =  24"l 
4.x-3y  =  ll] 

9. 

9x-07/  =  52| 
8?/  — 3ic=   8J 

1^3.      3x-^2y  =  32-\ 
20x  —  3y=    1} 

10. 

5x-3?/=   41 
122/-7a:  =  10j 

4.    llx  —  ly  =  3T\ 
8a:  +  92/  =  41j 

11. 

9y-7x  =  131 
ISx  — 77/=   9J 

5.      7a:+    6y  =  m^ 
13x-l\y  =  10i 

12. 

bx-2y=   511 
19a: -32/ =  180  J 

6.  6x  — 72/  =  421  13.    4.x-\-   92/  =  106\ 
7a;-62/  =  75j  8a;  +  172/  =  198  J 

7.  10ic+   93/  =  290\  14.      8ic  +  32/  =  31 
12a; -11 2/ =  130  J  12x-^9y  =  3} 


Elimination  by  Comparison. 


Solve:  2a;  — 92/  =  11 


3a:  — 42/ 


:"} 


2ic-9y  =  ll,  (1) 

3x-4y  =  7.  (2) 

Transpose  9  ?/  in  (1)  and  4  y  in  (2),       2  a;  =  11  +  9  y,  (3) 

3x=7  +  4y.  (4) 

Divide  (3)  by  2  and  (4)  by  3,  x=  ll±i^,  (S) 


74-4; 
3 


(6) 

11  +  9?/      7  +  4?/ 
Equate  the  values  of  x,  r —  =  — ^ — •  (7) 


166  ALGEBRA. 

Reduce  (7),  33  +  27  y  =  14  +  8  y, 

ldy=-  19. 
.•.y=-  1. 

•Substitute  the  value  of  y  in  (1),     2x4-9=11. 

.■.x=l. 

191.  Hence,  to  eliminate  an  unknown  number  by  com- 
parison, 

From  each  equation  obtain  the  value  of  one  of  the  unknown 
numbers  in  terms  of  the  other. 

Form  an  equation  from  these  equal  values  and  reduce  the 
equation. 

Note.     If,  in  the  last  example,  (3)  be  divided  by  (4),  the  resulting 

equation,  -  =  >  would,  when  reduced,  give  the  value  of  y. 

This  is  the  shortest  method,  and  therefore  to  be  preferred. 


Exercise  70. 
Solve  by  comparison  : 

j/  1.      ir  +  15?/  =  53\  8.    ^y  —  lx=   4\ 

j       2.    4a;+    9ij  =  51\  9.    21?/  +  20a^  =  165\ 

V          8x  — 13?/=   9j  772/-30x=:295j 

3.  4a;  +  3?/  =  48"l  10.    llx  —  10ij  =  U^ 
5y  —  3x  =  22j  5x+   7?/  =  41J   . 

4.  2x  +  Si/  =  4:S\  11.    7?/  — 3if  =  139\ 


2/  =  431 


lOa;-    ij=   7j  2x-]-5i/=   91j 

5.  5x—    7?/=   33\  12.    17^  +  12?/=    59\ 
lla:  +  127/  =  100j  19a^-   4?/  =  153j 

6.  5£c-f  7?/  =  43\  13.    24:r+    7y=    27  \ 
110!  + 9y=  69.1  8a;  — 33?/  =  115j 

7.  8a;  — 2l?/=   33\  14.    x=:3ij  —  19\ 
6x  -f  35.7  :=- 177  J  7j  =  3x~23i 


SIMULTANEOUS    EQUATIONS.  157 

192.   Each  equation  must   be    simplified,   if  necessary, 
before  the  elimination  is  performed. 

Solve:       (x-l)(!/-\-2)  =  (x-3)(l/-l)-j-S^        • 
2x-l      3(y-2)^  I 

5  4  J 

{X  -l){y  +  2)={x-S){y-l)  +  S  (1) 

2x^l_3_fc^^^  (2) 

Simplify  (1),  xy  +  2x-y-2  =  xy-x  —  Sy  +  3  +  8. 

Transpose  and  combine,  Sx  +  2y  =  IS.  (3) 

■      Simplify  (2),  8 x  -  4  —  15  y  +  30  =  20. 

Transpose  and  combine,  Sx  —  16y  =  —6.  (4) 

Multiply  (3)  by  8,  2ix+16y=  104.  (6) 

Multiply  (4)  by  3,  24  x  —  45  ?/  =  -  18.  (6) 

Subtract  (6)  from  (5),  Qly=  122. 

.-.  y  =  2. 
Substitute  the  value  of  y  in  (3),    3x4-4  =  13. 

.-.  X  =  3. 


Exercise  71. 


Solve  : 


1.    x(j/-\-7)  =  y(x  +  l) 


;"} 


2  3 


2a^  +  20  =  3y+l      J  ^  +  3      y  — 2 

5(a:  +  3)  =  3(2/-2)  +  2. 

5.    (x  +  l)(y  +  2)-(x  +  2)(>,+  l)  =  -l\ 
3(^  +  3)-4(y+4)  =  -8  J 

^    x  —  2      10  — K      y  — 10    ^ 

2y  +  4_2£-f-j/^.H-13f 
3  8  4      •' 


158 


ALGEBRA. 


x  +  1      y  +  2^2(x-y)^ 
3  4  5 


x—3      y—3 


4  3 

Sx  —  2y  .   5x  —  37j 


2y  —  x 


+ 


^  +  1 


2x-3y      4.x-3y^ 
3       "^       2  ^^ 


16. 


11    -^ 

l-3x  _ll-3y 

7      "'      5       . 


2a^-y  +  3      x-2y^3_ 
^•3  4  -^ 

3cc  — 4?/  +  3  ,  4.x  — 2y  — 9_ 


10.    I^a^  =  li2/  +  4T\\ 

13  3 


17.    5:r  — i(5?/  +  2)  =  32 
37/+i(^  +  2)  =  9 


11. 


12. 


a:+2y+3  4ic— 5i/+6  |     18.    3cc— 0.25?/==28 

3  _  19  f  0.12a^+0.72/=2.54 

6ic— 5?/+4~3a;+2?/+l 


0?  4"  ?/ 15 

y  —  x~  8 

32/  +  44 


9a; 


=  100 


19.    7(x-l)  =  3(^  +  8) 

4a;  +  2_5y  +  9 
9      ~      2 


3a;  —  5?/  ,   ^      2a;  +  v"l 
13.    _^  +  3  =  -^1 

a;-2y_a;   ,   y 
^  4~-2  +  3      J 


20.    7a;  +  i(22/  +  4)  =  16^ 
32/-i(^  +  2)  =  8     J 


14. 


4a;  — 3y  — 7_3a;  2y      5 

5  ~  10  15      6 

y  — 1       a;      3y  ^y  — a;   ,   -^   .    1 
3     "^2       20  15    "^6  "^10  J 


21. 


SIMULTANEOUS    EQUATIONS. 
5x  —  61/ 


159 


13 


-\-Sx  =  4.2/  —  2 


Bx-^-Si/      3x  —  2y 

6  4        ~ 

^^     5£c  — 3      3a;  — 19       , 
22.    — ;: =  4 


2y-2 
3?/  —  X 


2  2*3 

2a;  +  y      9a;  — 7_3(3/  +  3)       4a;  +  5y 


8 


16 


23.    3y  +  ll  =  i^^^^^^±^  +  31-4a. 
a;  — 2/H-4 

(^  +  7)(2/-2)  +  3  =  2x^/-(y-l)(.T  +  l) 


8y  +  7      6x-3y^         4y-9 
10     "^  2y  —  8  ■*"       5 


Literal  Simultaneous  Equations. 

193.   The  method  of  solving  literal  simultaneous  equa- 
tions is  as  follows : 


Solve : 


ax- -j- bij  =  m ") 
cx-\-di/=  n  J 


Multiply  (1)  by  c, 
Multiply  (2)  by  a, 
Subtract  (4)  from  (3), 

Divide  by  coefficient  of  y, 


CX  +  dy  =  n 
acx-\-  bey  =  cm 

acx  +  ady=an 

(6c  —  ad)y  =  cm  —  an 

_  cm  —  an 

^~  hc-ad 


(1) 
(2) 
(3) 
(4) 


160 


ALGEBRA. 


To  find  the  value  of  x : 

Multiply  (1)  by  d, 
Multiply  (2)  by  6, 
Subtract  (6)  from  (6), 

Divide  by  coefficient  of  x, 


adx  +  hdy  =  dm 

hex  +  bdy  =  bn 

{ad  —hc)x  —  dm  —  bn 
dm  —  bn 
ad  — be 


x  = 


(5) 

(6) 


Exercise  72. 


Solve : 


1.    x-\-y=a'\        3.    7nx -\- 717/ =  a ^ 
x  —  y  =  b)  px-\-'qij  =  h   J 


5.    mx  —  ny^^r 


X  —  ny^=r    "i 
'x  -\-  n'y  =  r'  ) 


2.    ax-^-by^cy    4.    ax-{-hy  =  e 
J>x-\-qy  =  r  j  ax-\-cy  =  d 


X   ,   y 

7.    ~  +  f  =  c 

a       0 


X  71 

0       a 


12. 


6.    ax-\-hy  =  G'\ 
dx-\-fy=c^} 

X  —  V  + 1 
x—y—1  [ 

x-^y-\-l 
^  +  ^  — 1 


=  h 


8.    ahx  +  cdy  =  2 

d  —  b 
ax  —  ci/^——z— 
^         bd 


9. 


b-\-y      ^a-\-x 
ax-\-2by  =  d 


13.    ax  =  by-\- 


+  h' 


14. 


2 
(a  —  b)  X  =  (a -\- b)  y 

ax-\-by  =  c^  1 

a  ^  .  I 


b-j-y       a-\-x 


10. 


X 


y 


a-\-b       a  —  b 
^      I      2/ 


1 

a-\-b 
1 


a-\-  b   '   a  —  b      a  —  b 

a  (a  —  x^- 

a(y  —  b  —  x)  =b  (y  —  b) 


15. 


11.    a(a  —  x)—b(x-]-y-'a)^      16. 


a-\-  b       a  —  b 

X  —  y x-\-y 

2ab   ~a^-i-b^ 

bx  —  bc  =  ay  —  ac^ 
X  —  y=-ci  —  b         J 


SIMULTANEOUS    EQUATIONS.  161 


17.    ^-'^ 


y-b  I 

a{x  —  (i)-\-h{y  —  5) -|  ahc  =  0  J 

18.  (a-fi)a;  — (a~J)2/  =  4a^  -i 
(a  — ^.)a;  +  (a  +  6)y  =  2a2  — 252I 

19.  (x^a){y^h)---{x--a){y--h)=^2{a--'hy 
x  —  y-i-2(a  —  b)  =  0 

20.  (a-i-b)(x  +  y)-(a-b)(x-y)  =  a'^ 
(a-b)(x  +  y)  +  Ca  +  b)(x-y)  =  b'  J 


194.  Fractional  simultaneous  equations,  of  which  the 
denominators  are  simple  expressions  and  contain  the 
unknown  numbers,  may  be  solved  as  follows: 


1)    Solve; 

a  ,   b          1 
X      y 
c  ,   d 

x^y          J 

■ 

a  ,  b 

-  +  -=  m. 

X      y 

c  ,  d 

-  +  -  =  n. 

X      y 

Multiply  (1)  hj  c, 

ac  ,  be 

h  —  =  cm. 

X       y 

Multiply  (2)  by  a, 

ac  ,  ad 

— 1 =  an. 

X       y 

Subtract  (4)  from 

(3), 

be-ad 
=  cm- 

V 

-an. 

Multiply  both  sides  by  y,    bc  —  ad=  {cm 

-  an)  y. 

bc- 
'-'^^  cm- 

■ad 
-an 

Multiply  (1)  by  d, 

ad  ,  bd      ^ 

—  H =  dm. 

X       y 

Multiply  (2)  by  6, 

be     bd_ 
X       y  ~ 

=  bn. 

(1) 

(2) 
(3) 
(4) 


(5) 
<6) 


162 


ALGEBRA. 


Subtract  (6)  from  (5), 


ad  — be 


^dm  —  bn. 


Itiply  both  sides  by  x,    ud  —  bc=  {d 

m  —  bn)  X. 

ad  — be 
dm  —  bn 

Solve:                 |-  +  A=7    ^ 
3a:       5y 

^          ^    -3 

6x      lOy         J 

5    ,     2 

3^  +  5^-'- 
6x      lOy 


Multiply  (2)  by  4, 
Add  (1)  and  (3), 
Divide  both  sides  by  19, 


U_2_ 
Sx      by 

3a; 


=  12. 
=  19. 


Substitute  the  value  of  x  in  (1),  5  +  7-  =  7. 

5y 


Transpose, 

Divide  both  sides  by  2, 


2_ 
oy 

2_ 
by 
J_ 
by 


=  2. 

=  1. 


(1) 

(2) 
(8) 


1. 


Solve : 

1      2 

-  +  -=10 

X       y 

X      y 

X       y 

X      y 


Exercise  73. 


2        5        4  -^ 

X      3y      27 

5. =  5 

X       y 

11       11 

^x'^y-72j 

^-5  =  6 

^     y 

X       y 

a   ,    h       ac^ 
6.    -  +  -  =  — 
X       y       b 

X      y         J 

b  ^.a  be 
X      y~  a  . 

SIMULTANEOUS    EQUATIONS. 


163 


,.2,3      ^^ 

7. hr=^  I 

ax      by         I 

ax      by 


1 =  7?t  +  7i 

nx      my 

n      m          ,  ,     2 

X       y 


a   .    h  \ 

=  7)1 

X      y 


195.  If  three  simultaneous  equations  are  given,  involv- 
ing three  unknown  numbers,  one  of  the  unknown  numbers 
must  be  eliminated  between  two  pairs  of  the  equations; 
then  a  second  between  the  resulting  equations. 

196.  Likewise,  if  four  or  more  equations  are  given,  in- 
volving four  or  more  unknown  numbers,  one  of  the 
unknown  numbers  must  be  eliminated  between  three  or 
more  pairs  of  the  equations;  then  a  second  between  the 
pairs  that  can  be  formed  of  the  resulting  equations ;  and 
so  on. 


Solve : 


2x  —  Sy-{'4:Z=  4 
Sx  +  5y—7z  =  12 
5x  —  y    — Sz=    5 


Eliminate  z  between  two  pairs  of  these  equations. 


Multiply  (1)  by  2, 

(3)  is 

Add, 

Multiply  (1)  by  7, 
Multiply  (2)  by  4, 
Add, 

Multiply  (6)  by  7, 

(6)  is 

Subtract  (5)  from  (7), 


ix  —  Gy-^Sz=    8 


5x—    y- 

-8z 

=    5 

9x-7y 

=  13 

14 

x-2ly  +  2Sz 

=  28 

12 

x  +  20y'- 

28  z 

=  48 

26 

X-      y 

=  76 

182x- 

-7y 

=  532 

9x- 

-iy 

=    13 

173  X 


(1) 
(2) 
(3) 


(4) 
(6) 

(6) 
(7) 


Substitute  the  value  of  x  in  (6), 
Substitute  the  values  of  x  and  y  in  (1),  6 


=  519 

.-.  X 

=  3. 

78- 

y  = 

76. 

.•. 

y  = 

2. 

),  6- 

-6  +  4z  = 

4. 

.• 

z  = 

1. 

164 


ALGEBRA. 


^ 


Exercise   74. 
Solve : 

1.    5x-}-3i/  —  6z  =  4:^  10.    Sx  —  y  +  z  =  lT 

3x~i/  +  2z  =  8     I  5x-\-3i/  —  2z  =  10 

x~2y-\-2z  =  2     J  7x-^4.i/  —  5r^  =  3 


2.  4:X  —  5i/-}-2z  =  6  ^  11.  x-^?/-{-z  =  5  ^ 
2x-{-3f/  —  z  =  20  I  3x  —  5i/-{-7z  =  75i 
7x  —  4:j/-^3z  =  35j  9^-11^  + 10  =  0  J 

3.  x  +  i/-i-z  =  6  -)  12.  x  +  2ij-\-3z  =  6  -) 
5x-\-4i/-\-3z  =  22  I  2x-\-4.ij-{-2z  =  S  i 
15x-\-10i/-}-Qz  =  53}  3x  +  22j-i-8z  =  101) 

4.  Ax  —  3i/  +  z  =  9    1  13.    a;  — 3^  — 2^  =  1 

a;  — 47/  +  3^=:2    J  5x-{-2ij  — z  =  12 

5.  8:r  +  4^  — 3^  =  6^  14.  3ic  — 2?/  =  5  ^ 
x  +  3ij  —  z  =  7  I  4ic  — 3?/+2^  =  ll  I 
4:x  —  5ij  +  4.z  =  s\  x  —  2y  —  5z  =  —  7} 

6.  12x  +  5?/  — 4^  =  29^ 

17x  —  j/  —  z  =  15       J 

7.  2/  — a;  +  ^  =  — 5  ^  16.  2a;  — 3^  =  3^ 
^  — ?/  — x  =  — 25  I  3y  —  4:z  =  7  I 
ic  +  7/  +  ^  =  35      J  4s  — 5a;  =  2j 

8.  x-i-ij  +  z  =  30            •)  17.    3a;  — 4?/  +  6s  =  l 
Sx  +  4:y-]-2z  =  50     I  2a3  +  2?/  — s  =  l 
27a;  +  92/  +  3;s  =  64j  7a;  — 6?/  + 7^  =  2 

9.  157/  =  24s  — 10a;  +  41  ^  18.  7a;  — 3?/  =  30  ^ 
15a;  =  122/  — 16^  +  10  I  9y  —  5z  =  34.  I 
18  a;  — (7«-13)  =  14yJ  x  +  y-\-z  =  33) 


SIMULTANEOUS    EQUATIONS. 


165 


19.   a.  +  |  +  |  =  6 

.    z       X 

2^+2+3  =  - 
.+  1  +  1  =  17 


20.    -  +  -  =  5 
X      y 

y       z 
2-1  =  5 

Z         X 


21.    i  +  i-i=a 
X      y      z 

a;      ?/       ^ 
1,1       1 


23.    ?-f +  ^=7i 
X       by       z         ^ 

r^  +  ;^  +  -  =  10i 

3a;      2»/      s  * 


4         14 

24. 

?-?  +  l  =  2.9 

a:       2/       ;s 

9+12_«  =  i4.9 
y       z        X 

25. 

a:      2/      « 

2-2-2=0 

«      2/ 

i+i-i=o 

a;   ^      3             J 

22.    hz-\-cy  =  a  '\ 
az-\-cx  =  b    If 
ay-{-bx  =  c  J 


26.  aa;  +  J?/  4"  ^^  =  ^ 
aa;  —  by  —  cz  =  b 
ax -\- cy -\- bz  =  c 


27. 


2x  —  y Sy-\-2z x  —  y  —  z 


28. 


x—y ?/  —  z x-\-z X  —  a  —  b 

a  b  c  a-^b-\-c 


*  In  example  21,  subtract  from  the  sum  of  the  three  equations  each 
equation  separately. 

t  In  example  22,  multiply  the  equations  by  a,  6,  and  c,  respectively, 
and  from  the  sum  of  the  results  subtract  the  double  of  each  equation 
separately. 


CHAPTER   XII. 

Problems  producing  Simultaneous  Equations. 

197.  It  is  often  necessary  in  the  solution  of  problems 
to  employ  two  or  more  letters  to  represent  the  numbers  to 
be  found.  In  all  cases  the  conditions  must  be  sufficient  to 
give  just  as  many  equations  as  there  are  unknown  numbers 
employed. 

If  there  are  more  equations  than  unknown  numbers, 
some  of  them  are  superfluous  or  contradictory  ;  if  there 
are  less  equations  than  unknown  numbers,  the  problem  is 
indeterminate  or  impossible. 

(1)  When  the  greater  of  two  numbers  is  divided  by  the 
less  the  quotient  is  4  and  the  remainder  3 ;  and 
when  the  sum  of  the  two  numbers  is  increased  by 
38,  and  the  result  divided  by  the  greater  of  the 
two  numbers,  the  quotient  is  2  and  the  remainder  2. 
Eind  the  numbers. 


Let 

X  =  the  greater  number, 

and 

y  =  the  smaller  number. 

Then 

^-^  =  4, 

X  +  y  +  38  -  2      ^ 
and =  2. 

X 

From  the  solution  of  these  equations,  x  =  47,  and  y  =  11. 

(2)  If  A  give  B  $10,  B  will  have  three  times  as  much 
money  as  A.  If  B  give  A  $10,  A  will  have  twice 
as  much  money  as  B.     How  much  has  each  ? 


PROBLEMS.  167 

Let  X  =  the  number  of  dollars  A  has, 

and  y  =  the  number  of  dollars  B  has. 

Then  y+  10  =  the  number  of  dollars  B  has,  and  x  —  10  =  the 
number  of  dollars  A  has  after  A  gives  $10  to  B. 

.:y+10  =  S{x  —  10),  and  a:  +  10  =  2  (y  —  10). 

From  the  solution  of  these  equations,  x  =  22,  and  y  =  26. 

Therefore,  A  has  $22  and  B  §26. 

Exercise  75. 

1.  The  sum  of  two  numbers  divided  by  2  gives  as  a  quo- 

tient 24,  and  the  difference  between  them  divided  by 
2  gives  as  a  quotient  17.     What  are  the  numbers  ? 

2.  The  number  144  is  divided  into  three  numbers.     When 

the  first  is  divided  by  the  second,  the  quotient  is  3 
and  the  remainder  2  ;  and  when  the  third  is  divided 
by  the  sum  of  the  other  two  numbers,  the  quotient 
is  2  and  the  remainder  6.     Find  the  numbers. 

3.  Three  times  the  greater  of  two  numbers  exceeds  twice 

the  less  by  10 ;  and  twice  the  greater  together  with 
three  times  the  less  is  24.     Find  the  numbers. 

4.  If  the  smaller  of  two  numbers  is  divided  by  the  greater, 

the  quotient  is  0.21  and  the  remainder  0.0057 ;  but 
if  the  greater  is  divided  by  the  smaller,  the  quo- 
tient is  4  and  the  remainder  0.742.  What  are  the 
numbers  ? 

5.  Seven  years  ago  the  age  of  a  father  was  four  times  that 

of  his  son ;  seven  years  hence  the  age  of  the  father 
will  be  double  that  of  the  son.     What  are  their  ages  ? 

6.  The  sum  of  the  ages  of  a  father  and  son  is  half  what 

it  will  be  in  25  years ;  the  difference  between  their 
ages  is  one-third  of  what  the  sum  will  be  in  20  years. 
What  are  their  ages  ? 


168  ALGEBRA. 

7.  If  B  give  A  $25,  they  will  have  equal  sums  of  money ; 

but  if  A  give  B  $22,  B's  money  will  be  double  that 
of  A.     How  much  has  each  ? 

8.  A  farmer  sold  to  one  person  30  bushels  of  wheat  and 

40  bushels  of  barley  for  $67.50 ;  to  another  person 
he  sold  50  bushels  of  wheat  and  30  bushels  of  barley 
for  $S5.  What  was  the  price  of  the  wheat  and  of 
the  barley  per  bushel  ? 

9.  If  A  give  B  $5,  he  will  then  have  $6  less  than  B  ;  but 

if  he  receive  $5  from  B,  three  times  his  money  will 
be  $20  more  than  four  times  B's.  How  much  has 
each? 

10.  The  cost  of  12  horses  and  14  cows  is  $1900 ;   the  cost 

of  5  horses  and  3  cows  is  $650.  What  is  the  cost 
of  a  horse  and  a  cow  respectively  ? 

Note  I.     A  fraction  of  which  the  terms  are  unknown  may  be 

represented  by  -• 

Ex.  A  certain  fraction  becomes  equal  to  ^  if  3  is  added  to 
its  numerator,  and  equal  to  f  if  3  is  added  to  its 
denominator.     Determine  the  fraction. 

Let  -  =  the  required  fraction. 

y 

By  the  conditions  =  h 

y 

and  -£-  =  f . 

From  the  solution  of  these  equations  it  is  found  that 
x  =  6, 
2/ =18. 
Therefore  the  fraction  =  y«^. 

11.  A  certain  fraction  becomes  equal  to  2  when  7  is  added 

to  its  numerator,  and  equal  to  1  when  1  is  subtracted 
from  its  denominator.     Determine  the  fraction. 


PROBLEMS.  169 

12.  A  certain  fraction  becomes  equal  to  ^  when  7  is  added 

to  its  denominator,  and  equal  to  2  when  13  is  added 
to  its  numerator.     Determine  the  fraction. 

13.  A   certain   fraction    becomes    equal   to   ^   when   the 

denominator  is  increased  by  4,  and  equal  to  |^ 
when  the  numerator  is  diminished  by  15.  Deter- 
mine the  fraction. 

14.  A  certain  fraction  becomes  equal  to  §  if  7  is  added  to 

the  numerator,  and  equal  to  |  if  7  is  subtracted 
from  the  denominator.     Determine  the  fraction. 

15.  Find  two  fractions  with  numerators  2  and  5  respec- 

tively, such  that  their  sum  is  1^;  and  if  their 
demominators  are  interchanged  their  sum  is  2. 

16.  A  fraction  which  is  equal  to  §  is  increased  to  y\  when 

a  certain  number  is  added  to  both  its  numerator 
and  denominator,  and  is  diminished  to  |  when  one 
more  than  the  same  number  is  subtracted  from 
each.      Determine  the  fraction. 

Note  II.  A  number  consisting  of  two  digits  which  are  unknown 
may  be  represented  by  10  a;  +  ?/,  in  which  x  and  y  represent  the  digits 
of  the  number.  Likewise,  a  number  consisting  of  three  digits  which 
are  unknown  may  be  represented  by  100  x-\-  \0y  +  z,m  which  x,  y, 
and  z  represent  the  digits  of  the  number. 

For  example,  consider  any  number  expressed  by  three  digits,  as 
364.  The  expression  364  means  300  +  00  +  4  ;  or,  100  times  3+10 
times  6  +  4. 

Ex.  The  sum  of  the  two  digits  of  a  number  is  8,  and  if 
36  be  added  to  the  number  the  digits  will  be 
interchanged.      What  is  the  number  ? 

Let  X  =  the  digit  in  the  tens'  place, 

and  y  =  the  digit  in  the  units'  place. 

Then  10  x  +  y  =  the  number. 

By  the  conditions,  x  +  y  =  8,  (1) 

and  10x  +  ?/  +  36=102/  +  x.  (2) 


170 


ALGEBRA. 

From  (2), 

9x- 

-9y  =  -36. 

Divide  by  9, 

X 

—  2/  =:  —  4. 

Add  (1)  and  (3), 

2ic  =  4. 
.-.  x=2. 

Subtract  (3)  from 

(1), 

2y=  12. 
.-.2/ =  6. 

Hence,  the  number  is 

I  26. 

17.  The  sum  of  the  two  digits  of  a  number  is  10,  and  if 

54  be  added  to  the  number  the  digits  will  be 
interchanged.      What  is  the  number  ? 

18.  The  sum  of  the  two  digits  of  a  number  is  6,  and  if 

the  number  be  divided  by  the  sum  of  the  digits 
the  quotient  will  be  4.      What  is  the  number  ? 

19.  A  certain  number  is  expressed  by  two  digits,  of  which 

the  first  is  the  greater.  If  the  number  is  divided  by 
the  sum  of  its  digits  the  quotient  is  7 ;  if  the  digits 
are  interchanged,  and  the  resulting  number  dimin- 
ished by  12  is  divided  by  the  difference  between  the 
two  digits,  the  quotient  is  9.     What  is  the  number  ? 

20.  If  a  certain  number  is  divided  by  the  sum  of  its  two 

digits  the  quotient  is  6  and  the  remainder  3  ;  if  the 
digits  are  interchanged,  and  the  resulting  number  is 
divided  by  the  sum  of  the  digits,  the  quotient  is  4 
and  the  remainder  9.     What  is  the  number  ? 

21.  If  a  certain  number  is  divided  by  the  sum  of  its  two 

digits  diminished  by  2,  the  qiTotient  is  5  and  the 
remainder  1 ;  if  the  digits  are  interchanged,  and  the 
resulting  number  is  divided  by  the  sum  of  the  digits 
increased  by  2,  the  quotient  is  5  and  the  remainder 
8.     Find  the  number. 

22.  The  first  of  the  two  digits  of  a  number  is,  when  doubled, 

3  more  than  the  second,  and  the  number  itself  is  less 
by  6  than  five  times  the  sum  of  the  digits.  What  is 
the  number  ? 


PROBLEMS.  171 

23.  A  number  is  expressed  by  three  digits,  of  which  the 

first  and  last  are  alike.  By  interchanging  the  digits 
in  the  units'  and  tens'  places  the  number  is  increased 
by  54  ;  but  if  the  digits  in  the  tens'  and  hundreds' 
places  are  interchanged,  9  must  be  added  to  four 
times  the  resulting  number  to  make  it  equal  to  the 
original  number.     What  is  the  number  ? 

24.  A  number  is  expressed  by  three  digits.     The  sum  of 

the  digits  is  21 ;  the  sum  of  the  first  and  second 
exceeds  the  third  by  3 ;  and  if  198  be  added  to  the 
number,  the  digits  in  the  units'  and  hundreds'  places 
will  be  interchanged.     Find  the  number. 

25.  A  number  is  expressed  by  three  digits.     The  sum  of 

the  digits  is  9 ;  the  number  is  equal  to  forty-two 
times  the  sum  of  the  first  and  second  digits ;  and 
the  third  digit  is  twice  the  sum  of  the  other  two. 
Find  the  number. 

26.  A  certain  number,  expressed  by  three  digits,  is  equal 

to  forty-eight  times  the  sum  of  its  digits.  If  198 
be  subtracted  from  the  number,  the  digits  in  the 
units'  and  hundreds'  places  will  be  interchanged ; 
and  the  sum  of  the  extreme  digits  is  equal  to  twice 
the  middle  digit.     Find  the  number. 

Note  III.  If  a  boat  moves  at  the  rate  of  x  miles  an  hour  in  still 
water,  and  if  it  is  on  a  stream  that  runs  at  the  rate  of  y  miles  an  hour, 
then 

x+  y  represents  its  rate  down  the  stream, 
X  —  y  represents  its  rate  up  the  stream. 

27.  A  waterman  rows  30  miles  and  back  in  12  hours.     He 

finds  that  he  can  row  5  miles  with  the  stream  in  the 
same  time  as  3  against  it.  Find  the  time  he  was 
rowing  up  and  down  respectively. 


172  ALGEBRA. 

28.  A  crew  which  can  pull  at  the  rate  of  12  miles  an  hour 

down  the  stream,  finds  that  it  takes  twice  as  long  to 
come  up  the  river  as  to  go  down.  At  what  rate  does 
the  stream  flow  ? 

29.  A  man  sculls  down  a  stream,  which  runs  at  the  rate  of 

4  miles  an  hour,  for  a  certain  distance  in  1  hour  and 
40  minutes.  In  returning  it  takes  him  4  hours  and 
15  minutes  to  arrive  at  a  point  3  miles  short  of  his 
starting-place.  Find  the  distance  he  pulled  down 
the  stream  and  the  rate  of  his  pulling. 

30.  A  person  rows  down  a  stream  a  distance  of  20  miles 

and  back  again  in  10  hours.  He  finds  he  can  row 
2  miles  against  the  stream  in  the  same  time  he  can 
row  3  miles  with  it.  Find  the  time  of  his  rowing 
down  and  of  his  rowing  up  the  stream  ;  and  also  the 
rate  of  the  stream. 

Note  IV.  When  commodities  are  mixed,  the  quantity  of  the 
mixture  is  equal  to  the  quantity  of  the  ingredients ;  the  cost  of  the 
mixture  is  equal  to  the  cost  of  the  ingredients. 

Ex.  A  wine-merchant  has  two  kinds  of  wine  which  cost 
72  cents  and  40  cents  a  quart  respectively.  How 
much  of  each  must  he  take  to  make  a  mixture  of 
50  quarts  worth  60  cents  a  quart  ? 

Let  X  =  the  required  number  of  quarts  worth  72  cents  a 

quart, 
and  y  =  the  required  number  of  quarts  worth  40  cents  a 

quart. 
Then,    72  x  =  the  cost  in  cents  of  the  first  kind, 

40  y=  the  cost  in  cents  of  the  second  kind  of  wine, 
and  3000  =  the  cost  in  cents  of  the  mixture. 

.•.x  +  y  =  50, 
72x  +  40y  =  3000. 

Erom  which  equations  the  values  of  x  and  y  may  be  found. 


PROBLEMS.  173 

31.  A  grocer  mixed  tea  that  cost  him  42  cents  a  pound 

with  tea  that  cost  him  54  cents  a  pound.  He  had 
30  pounds  of  the  mixture,  and  by  selling  it  at  the 
rate  of  60  cents  a  pound,  he  gained  as  much  as  10 
pounds  of  the  cheaper  tea  cost  him.  How  many 
pounds  of  each  did  he  put  into  the  mixture  ? 

32.  A  grocer  mixes  tea  that  cost  him  90  cents  a  pound 

with  tea  that  cost  him  28  cents  a  pound.  The  cost 
of  the  mixture  is  $61.20.  He  sells  the  mixture  at 
50  cents  a  pound,  and  gains  $3.80.  How  many 
pounds  of  each  did  he  put  into  the  mixture  ? 

33.  A  farmer  has  28  bushels  of  barley  worth  84  cents  a 

bushel.  With  his  barley  he  wishes  to  mix  rye 
worth  $1.08  a  bushel,  and  wheat  worth  $1.44  a 
bushel,  so  that  the  mixture  may  be  100  bushels, 
and  be  worth  $1.20  a  bushel.  How  many  bushels 
of  rye  and  of  wheat  must  he  take  ? 

Note  V.  It  is  to  be  remembered  that  if  a  person  can  do  a  piece 
of  work  in  x  days,  the  part  of  the  work  he  can  do  in  one  day  will  be 
represented  by  ^. 

Ex.  A  and  B  together  can  do  a  piece  of  work  in  48  days  ; 
A  and  C  together  can  do  it  in  30  days  ;  B  and  C 
together  can  do  it  in  26f  days.  How  long  will  it 
take  each  to  do  the  work  ? 

Let  X  =  the  number  of  days  it  will  take  A  alone  to  do  the  work, 

y  =  the  number  of  days  it  will  take  B  alone  to  do  the  work, 

and  z  =  the  number  of  days  it  will  take  C  alone  to  do  the  work. 

Then,  - »  -  »  - »  respectively,  will  denote  the  part  each  can  do 
in  a  day, 

and  -  +  -  will  denote  the  part  A  and  B  together  can  do  in  a  day; 
but  —  will  denote  the  part  A  and  B  together  can  do  in  a  day. 


174 


ALGEBRA. 

x^  y      48 

(1) 

1  +  1  =  1 
x^  z      30 

(2) 

1,1-    1    _  3 

y'^  z      26f      80 

(3) 

2      2       2      _  11 
X      y      z          120 

(4) 

2      2             _  1 
x'^  y                 24 

(5) 

Therefore, 

Likewise, 

and 

Add  (1),  (2),  and  (3), 

Multiply  (1)  by  2, 

■^      y 

2       1 
Subtract  (5)  from  (4),  ~—  7iR 

Z       isO 

.-.  z  =  40. 

2        1 

Subtract  the  double  of  (2)  from  (4),  -  = — 

'  y      40 

.-.  y  =  80. 

2       1 

Subtract  the  double  of  (3)  from  (4),  -  =  r- 
^  '  ^  '    X      60 

.-.  X  =  120. 

34.  A  and  B  together  earn  ^40  in  6  days ;  A  and  C  to- 

gether earn  ^54  in  9  days;  B  and  C  together  earn 
$80  in  15  days.     What  does  each  earn  a  day  ? 

35.  A  cistern  has  three  pipes,  A,  B,  and  C.     A  and  B  will 

fill  it  in  1  hour  and  10  minutes ;  A  and  C  in  1  hour 
and  24  minutes ;  B  and  C  in  2  hours  and  20  minutes. 
How  long  will  it  take  each  to  fill  it  ? 

36.  A  warehouse  will  hold  24  boxes  and  20  bales;  6  boxes 

and  14  bales  will  fill  half  of  it.  How  many  of  each 
alone  will  it  hold  ? 

37.  Two  workmen  together  complete  some  work  in  20  days; 

but  if  the  first  had  worked  twice  as  fast,  and  the 
second  half  as  fast,  they  would  have  finished  it  in 
15  days.     How  long  would  it  take  each  alone  to  do  - 
the  work  ? 

38.  A  purse  holds  19  crowns  and  6  guineas;  4  crowns  and 

5  guineas  fill  ^^  of  it.  How  many  of  each  alone 
will  it  hold  ? 


PROBLEMS.  175 

39.  A  piece  of  work  can  be  completed  by  A,  B,  and  C  to- 

gether in  10  days  ;  by  A  and  B  together  in  12  days; 
by  B  and  C,  if  B  works  15  days  and  C  30  days. 
How  long  will  it  take  each  alone  to  do  the  work  ? 

40.  A  cistern  has  three  pipes,  A,  B,  and  C.     A  and  B  will 

fill  it  in  a  minutes  ;  A  and  C  in  b  minutes  ;  B  and 
C  in  c  minutes.  How  long  will  it  take  each  alone 
to  fill  it  ? 

Note  VI.  In  considering  the  rate  of  increase  or  decrease  in  quan- 
tities, it  is  usual  to  take  100  as  a  common  standard  of  reference^  so  that 
the  increase  or  decrease  is  calculated  for  every  100,  and  therefore 
called  per  cent. 

It  is  to  be  observed  that  the  representative  of  the  number  resulting 
after  an  increase  has  taken  place  is  100  -1-  increase  per  cent;  and 
after  a  decrease,  100  —  decrease  per  cent. 

Interest  depends  upon  the  time  for  which  the  money  is  lent,  as  well 
as  upon  the  rate  charged ;  the  rate  charged  being  the  rate  on  the 
principal  for  one  year.     Hence, 

,    .  ,       ,       Principal  X  Rate  X  Time 
Simple  mterest  = — r » 

where  Time  means  number  of  years  or  fraction  of  a  year. 
Amount  =  Principal  +  Interest. 

In  questions  relating  to  stocks,  100  is  taken  as  the  representative 
of  the  stocky  the  price  represents  its  market  value,  and  the  per  cent 
represents  the  interest  which  the  stock  bears.  Thus,  if  six  per  cent 
stocks  are  quoted  at  108,  the  meaning  is,  that  the  price  of  §100  of  the 
stock  is  $108,  and  that  the  interest  derived  from  $100  of  the  stock  will 
be  yf  ^  of  §100,  that  is,  $6  a  year.  The  rate  of  mterest  on  the  money 
invested  will  be  |§|  of  6  per  cent. 

41.  A  man  has  $10,000  invested.     For  a  part  of  this  sum 

he  receives  5  per  cent  interest,  and  for  the  rest  4  per 
cent ;  the  income  for  his  5  per  cent  investment  is 
$50  more  than  from  his  4  per  cent.  How  much 
has  he  in  each  investment  ? 


176  ALGEBRA. 

42.  A  sum  of  money,  at  simple  interest,  amounted  in  6 

years  to  §26,000,  and  in  10  years  to  130,000.  Find 
the  sum  and  the  rate  of  interest. 

43.  A  sum  of  money,  at  simple  interest,  amounted  in  10 

months  to  §26,250,  and  in  18  months  to  $27,250. 
Find  the  sum  and  the  rate  of  interest. 

44.  A  sum  of  money,  at  simple  interest,  amounted  in  m 

years  to  a  dollars,  and  in  7i  years  to  h  dollars.  Find 
the  sum  and  the  rate  of  interest. 

45.  A  sum  of  money,  at  simple  interest,  amounted  in  a 

months  to  c  dollars,  and  in  h  months  to  d  dollars. 
Find  the  sum  and  the  rate  of  interest. 

46.  A  person  has  a  certain  capital  invested  at  a  certain 

rate  per  cent.  Another  person  has  §1000  more 
capital,  and  his  capital  invested  at  one  per  cent 
better  than  the  first,  and  receives  an  income  §80 
greater.  A  third  person  has  §1500  more  capital, 
and  his  capital  invested  at  two  per  cent  better 
than  the  first,  and  receives  an  income  §150  greater. 
Find  the  capital  of  each,  and  the  rate  at  which 
it  is  invested. 

47.  A  person  has  §12,750  to  invest.      He  can  buy  three 

per  cent  bonds  at  81,  and  five  per  cents  at  120. 
Find  the  amount  of  money  he  must  invest  in  each  in 
order  to  have  the  same  income  from  each  investment. 

48.  A  and  B  each  invested  §1500  in  bonds  ;  A  in  three 

per  cents  and  B  in  four  per  cents.  The  bonds  were 
bought  at  such  prices  that  B  received  $5  interest 
more  than  A.  After  both  classes  of  bonds  rose  10 
points,  they  sold  out,  and  A  received  §50  more  than 
B.     What  price  was  paid  for  each  class  of  bonds  ? 


PROBLEMS.  177 

49.  A  person  invests  $10,000  in  three  per  cent  bonds, 

^16,500  in  three  and  one-half  per  cents,  and  has  an 
income  from  both  investments  of  $1056.25.  If  his 
investments  had  been  $2750  more  in  the  three  per 
cents,  and  less  in  the  three  and  one-half  per  cents, 
his  income  would  have  been  62^  cents  greater. 
What  price  was  paid  for  each  class  of  bonds  ? 

50.  The  sum  of  $2500  was  divided  into  two  unequal  parts 

and  invested,  the  smaller  part  at  two  per  cent  more 
than  the  larger.  The  rate  of  interest  on  the  larger 
sum  was  afterwards  increased  by  1,  and  that  of  the 
smaller  sum  diminished  by  1 ;  and  thus  the  interest 
of  the  whole  was  increased  by  one-fourth  of  its 
value.  If  the  interest  of  the  larger  sum  had  been 
so  increased,  and  no  change  had  been  made  in  the 
interest  of  the  smaller  sum,  the  interest  of  the 
whole  would  have  been  increased  one-third  of  its 
value.  Find  the  sums  invested,  and  the  rate  per 
cent  of  each. 

Note  VII.  If  x  represents  the  number  of  linear  units  in  the  length, 
and  y  in  the  width,  of  a  rectangle,  xy  will  represent  the  number  of  its 
units  of  surface  ;  the  surface  unit  having  the  same  name  as  the  linear 
unit  of  its  sides. 

61.  If  the  sides  of  a  rectangular  field  were  each  increased 
by  2  yards,  the  area  would  be  increased  by  220 
square  yards  j  if  the  length  were  increased  and  the 
breadth  were  diminished  each  by  5  yards,  the  area 
would  be  diminished  by  185  square  yards.  What 
is  its  area  ? 

52.  If  a  given  rectangular  floor  had  been  3  feet  longer  and 
2  feet  broader  it  would  have  contained  64  square 
feet  more ;  but  if  it  had  been  2  feet  longer  and  3 
feet  broader  it  would  have  contained  68  square  feet 
more.     Find  the  length  and  breadth  of  the  floor. 


178  ALGEBRA. 

53.  In  a  certain  rectangular  garden  there  is  a  strawberry- 

bed  whose  sides  are  one-third  of  the  lengths  of  the 
corresponding  sides  of  the  garden.  The  perimeter  of 
the  garden  exceeds  that  of  the  bed  by  200  yards  ; 
and  if  the  greater  side  of  the  garden  be  increased 
by  3,  and  the  other  by  5  yards,  the  garden  will  be 
enlarged  by  645  square  yards.  Find  the  length 
and  breadth  of  the  garden. 

Note  VIII.     Care  must  be  taken  to  express  the  conditions  of  a 
problem  with  reference  to  the  same  principal  unit. 

Ex.  In  a  mile  race  A  gives  B  a  start  of  20  yards  and  beats 
him  by  30  seconds.  At  the  second  trial  A  gives  B 
a  start  of  32  seconds  and  beats  him  by  9j\  yards. 
Find  the  rate  per  hour  at  which  each  runs. 

Let  X  =  tlie  number  of  yards  A  runs  a  second, 
and     y  =  tlie  number  of  yards  B  runs  a  second. 
Since  there  are  1760  yards  in  a  mile, 

— -  =  the  number  of  seconds  it  takes  A  to 

run  a  mile, 

1740       ,  1750t-V      ,,  ^,        ,. 

and ^  =  the  number  of  seconds  B  was  runnmg 

^-nr.  in  the  first  and  second  trials, 

Hence, =  30,      respectively. 

y         3/ 

1750^^      1760      ^^ 

and        —  — =  32. 

y  X 

The  solution  of  these  equations  gives  x  =  5}f  and  y  =  5^^. 

Ki3  1 

That  is,  A  runs  7=^ '  or  r--- »  of  a  mile  in  one  second  : 
17dO  oUO 

and  in  one  hour,  or  3600  seconds,  runs  12  miles. 

Likewise,  B  runs  lOjYx  niiles  in  one  hour. 

54.  In  a  mile  race  A  gives  B  a  start  of  100  yards  and 

beats  him  by  15  seconds.  In  the  second  trial  A 
gives  B  a  start  of  45  seconds  and  is  beaten  by  22 
yards.     Find  the  rate  of  each  in  miles  per  hour. 


PROBLEMS.  179 

55.  In  a  mile  race  A  gives  B  a  start  of  44  yards  and  beats 

him  by  51  seconds.  In  the  second  trial  A  gives  B 
a  start  of  1  minute  and  15  seconds  and  is  beaten  by 
88  yards.     Find  the  rate  of  each  in  miles  per  hour. 

56.  The  time  which  an  express  train  takes  to  go  120  miles 

is  y^5  of  the  time  taken  by  an  accommodation-train. 
The  slower  train  loses  as  much  time  in  stopping  at 
different  stations  as  it  would  take  to  travel  20  miles 
without  stopping;  the  express-train  loses  only  half 
as  much  time  by  stopping  as  the  accommodation- 
train,  and  travels  15  miles  an  hour  faster.  Find 
the  rate  of  each  train  in  miles  per  hour. 

ST".  A  train  moves  from  P  towards  Q,  and  an  hour  later 
a  second  train  starts  from  Q  and  moves  towards  P 
at  a  rate  of  10  miles  an  hour  more  than  the  first 
.  train;  the  trains  meet  half-way  between  P  and  Q. 
If  the  train  from  P  had  started  an  hour  after  the 
train  from  Q  its  rate  must  have  been  increased  by 
28  miles  in  order  that  the  trains  should  meet  at  the 
half-way  point.     Find  the  distance  from  P  to  Q. 

58.  A  passenger-train,  after  travelling  an  hour,  meets  with 

an  accident  which  detains  it  one-half  an  hour ;  after 
which  it  proceeds  at  four-fifths  of  its  usual  rate;  and 
arrives  an  hour  and  a  quarter  late.  If  the  accident 
had  happened  30  miles  farther  on,  the  train  would 
have  been  only  an  hour  late.  Determine  the  usual 
rate  of  the  train. 

59.  A  passenger-train,  after  travelling  an  hour,  is  detained 

15  minutes  ;  after  which  it  proceeds  at  three-fourths 
of  its  former  rate,  and  arrives  24  minutes  late.  If 
the  detention  had  taken  place  5  miles  farther  on, 
the  train  would  have  been  only  21  minutes  late. 
Determine  the  usual  rate  of  the  train. 


180  ALGEBRA. 

60.  A  man  bought  10  oxen,  120  sheep,   and  46  lambs. 

The  cost  of  3  sheep  was  equal  to  that  of  5  lambs ; 
an  ox,  a  sheep,  and  a  lamb  together  cost  a  number 
of  dollars  less  by  57  than  the  whole  number  of 
animals  bought ;  and  the  whole  sum  spent  was 
$2341.50.  Find  the  price  of  an  ox,  a  sheep,  and 
a  lamb,  respectively. 

61.  A  farmer  sold  100  head  of  stock,  consisting  of  horses, 

oxen,  and  sheep,  so  that  the  whole  realized  f  11.75  a 
head ;  while  a  horse,  an  ox,  and  a  sheep  were  sold 
for  $110,  $62.50,  and  $7.50,  respectively.  Had  he 
sold  one-fourth  of  the  number  of  oxen  that  he  did, 
and  25  more  sheep,  he  would  have  received  the  same 
sum.  Find  the  number  of  horses,  oxen,  and  sheep, 
respectively,  which  were  sold. 

62.  A,  B,  and'  C  together  subscribed  $100.     If  A's  sub- 

scription had  been  one-tenth  less,  and  B's  one-tenth 
more,  C's  must  have  been  increased  by  $2  to  make 
up  the  sum ;  but  if  A's  had  been  one-eighth  more, 
and  B's  one-eighth  less,  C's  subscription  would  have 
been  $17.50.     What  did  each  subscribe? 

63.  A  gives  to  B  and  C  as  much  as  each  of  them  has ;   B 

gives  to  A  and  C  as  much  as  each  of  them  then  has ; 
and  C  gives  to  A  and  B  as  much  as  each  of  them 
then  has.  In  the  end  each  of  them  has  $6.  How 
much  had  each  at  first  ? 

64.  A  pays  to  B  and  C  as  much  as  each  of  them  has ;  B 

pays  to  A  and  C  one-half  as  much  as  each  of  them 
then  has ;  and  C  pays  to  A  and  B  one-third  of  what 
each  of  them  then  has.  In  the  end  A  finds  that  he 
has  $1.50,  B  $4.16§,  C  $0.58J.  How  much  had 
each  at  first  ? 


CHAPTEE  XIIT. 
Involution  and  Evolution. 

198.  The  operation  of  raising  an  expression  to  any 
required  power  is  called  Involution. 

Every  case  of  involution  is  merely  an  example  of  multi- 
plication j  in  which  the  factors  are  equal.     Thus, 

(2ay  =  2a^X2a''  =  Aa\ 

199.  A  power  of  a  simple  expression  is  found  by  multi- 
plying the  exponent  of  each  factor  by  the  exponent  of  the 
required  factor,  and  taking  the  product  of  the  resulting 
factors.  The  proof  of  the  law  of  exponents,  in  its  general 
form,  is  : 

(«"*)'*  =  «"»  X  «"•  X  a'"  X to  71  factors, 

— —  Qin  +  m  +  m+ to  n  terms. 

Hence,  if  the  exponent  of  the  required  power  be  a  com- 
posite number,  it  may  be  resolved  into  prime  factors,  the 
power  denoted  by  one  of  these  factors  may  be  found,  and 
the  result  raised  to  a  power  denoted  by  another,  and  so  on. 
Thus,  the  fourth  power  may  be  obtained  by  taking  the 
second  power  of  the  second  power ;  the  sixth  by  taking 
the  second  power  of  the  third  power ;  the  eighth  by  taking 
the  second  power  of  the  second  power  of  the  second  power. 

200.  Erom  the  Law  of  Signs  in  multiplication  it  is 
evident  that, 

I.    All  even  powers  of  a  number  are  positive. 
II.    All  odd  powers  of  a  number  have  the  same  sign  as 
the  number  itself. 


182  ALGEBRA. 

Hence,  no  even  power  of  any  number  can  be  negative; 
and  of  two  compound  expressions  whose  terms  are  identical 
but  have  opposite  signs,  the  even  powers  are  the  same. 
Thus, 

(h  —  ay=\  —  {a  —  h)Y  =  {a  —  h)\ 

201.  A  method  has  been  given,  §  83,  of  finding,  without 
actual  multiplication,  the  powers  of  binomials  which  have 
the  form  (adtib). 

The  same  method  may  be  employed  when  the  terms  of  a 
binomial  have  coefficients  or  exponents. 

(1)  {a  —  hy  =  a^-Sa^b  +  3aP-b^ 

(2)  (5x'-2ff, 

=  (5  xy  -  3  (5  xy  (2  if)  +  3  (5  «^  (2  i/f  -  (2  y^, 
=  125  a^«  - 150  xy  +  60  xY  -Sy^. 

(3)  (a  —  by  =  a'  —  4:a%  +  6a%'  —  4.ab^  +  b\ 

(4)  (x'-hjy, 

=(xy-4.(xy(iy)-\-6{xy(iyy-Ax\i2jT+(iy)^ 

=  :,^-2xhj-\-lxy-\xY-\-i^y\ 

202.  In  like  manner,  a  polynomial  of  three  or  more 
terms  may  be  raised  to  any  power  by  enclosing  its  terms 
in  parentheses,  so  as  to  give  the  expression  the  form  of 
a  binomial.     Thus, 

(1)    {a-\-b-\-cy=^\a^(b-\-c)\', 

=  a^-\-Za^(b^c)-^'6a(b-\-cy-\-{b-{-cy, 
=  a3^3a2^  +  3a2c  +  3a52  +  6a&c 

■^?>ac'-\'b^-^U^c-^Zbc''-\-^. 


INVOLUTION    AND    EVOLUTION.  183 

(2)    (a^'-2x^-\-Sx-{-4:y, 

=  {x^-2xy-{-2(x^-2a^(3x-\-^)-}-(Sx  +  4:y, 

=  x^-4:X^+4:X*+6x*—4.x^—16x^-\-9x^+24.x-\-16, 

=x'-4:x'  +  10x'-4.x^-7x'-\-24:X  +  16. 

Exercise  76. 
Write  the  second  members  of  the  following  equations  : 

1.  (ay=  11.   (2a^cy=  21.   (—3a'b^cy  = 

2.  (xy=  12.   (—5ax^if)^=       22.   (—Sxify  = 

3.  (xhfy=         13.   (—7m^nxyy=  23.   (—oa''bxy  = 

•■(f)=  "(-IS)-  -(-S')  = 

•■(S)=  "■<-+■)■=   -(-=r)= 

6.  (x-^2y=  16.   (2a-  — «/=  26.  (1— a  — a2)2== 

7.  (x  —  2y=  17.   (3a'4-2«)«=  27.  (2— 3a:+4a;y= 

8.  (if  +  3)^=  18.   (2x  —  yy=  28.  (1  — 2x  +  x2)3  = 

9.  (l  +  2jr)*=  19.   (xhj  —  2xr/)^=  29.  (1  — a;  +  a;2)3  = 
10.  (2m  — 1)3=  20.   {ab  —  3y=  30.  (l+ic  +  x^y^ 

Evolution. 

203.  The  operation  of  finding  any  required  root  of  an 
expression  is  called  Evolution. 

Every  case  of  evolution  is  merely  an  example  of  factor- 
ing^ in  which  the  required  factors  are  all  equal.      Thus, 

the   square,   cube,   fourth roots    of   an   expression   are 

found  by  taking  one  of  the  two^  three,  four equal  factors 

of  the  expression. 


184  ALGEBRA. 

204.  The  symbol  which  denotes  that  a  square  root  is  to 
be  extracted  is  V;  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  written  above  to  indicate  the  root, 
thus,   ^,   ^,  etc.,  signifies  the  third  root,  fourth  root,  etc. 

205.  Since  the  cube  of  a^^=a^,  the  cube  root  of  a^  =  a^. 
Since  the  fourth  power  of  2  a^  =  2*a^,  the  fourth  root  of 

Since  the  square  of  abc  =  a^bh^,  the  square  root  of  a%^c^ 

=  abc. 

d'         ..  „  a*      a%^    ,-  ^    „  a^h^      ah 

bince  the  square  oi  —  =  -^-5,  the  square  root  01  -^r^  =  — 

icy      x^if  x^if      xy 

Hence,  the  root  of  a  simple  expression  is  found  by 
dividing  the  exponent  of  each  factor  hy  the  index  of  the 
rootf  and  taking  the  product  of  the  resulting  factors. 

206.  It  is  evident  from  §  200  that 

I.   Any  even  root  of  a  positive  number  will  have  the 
double  sign,  ± . 

II.    There  can  be  no  even  root  of  a  negative  number. 

III.  Any  odd  root  of  a  number  will  have  the  same  sign 
as  the  number. 

VI  fi  7.2  4.  ^      3 , 


^i6xy^_    2xy 

^  81a''  ~-   3a*' 

But  V—  x^  is  neither  +  ^  nor  —  x,  for  (-|-  xy  =  -f  x^,  and 
(—xy  =  -{-x\ 

The  indicated  even  root  of  a  negative  number  is  called 
an  impossible,  or  imaginary,  number. 


INVOLUTION    AND    EVOLUTION.  185 

207.  If  the  root  of  a  number  expressed  in  figures  is  not 
readily  detected,  it  may  be  found  by  resolving  the  number 
into  its  prime  factors.  Thus,  to  find  the  square  root  of 
3,415,104  : 


2« 

3415104 

2^ 

426888 

32 

53361 

7 

5929 

7 

847 

11 

121 

11 

3,415,104  =  2«  X  32  X  72  X  111 


•••  73,415,104  =  2^X3  X7X  11  =  1848. 
Simplify:  Exercise  77. 

2.     ^—172S c^d'^xy,  ^S375b^W^,  \^3m696?V. 


3.  V5336UVy^.-,   V— 343^'    V^297-' 

4.  V25^^W  +  -v'8a«^V—  \/81a*^»«c*  —  v'32^^W. 


5.  ^27  xY  X  V243?/V  X  Vl6a;V. 

When  a  =  l,  5  =  3,  x=-2,  y  =  6,  find  the  values  of: 

6.  4  V2^  —  -^ahxy  +  5  ^a%^xy. 


7.    2  a  V8ax  +  Z»  Vl262/  +  4a^>a;V^»icy. 


8.  Va^  +  2a^»  +  ^>2x  Va^  +  3 a^^  +  3 a^-^ _p  ^•3. 

9.  ^b^  —  Sb^a  +  3^>a2  —  »»  -f-  V&'  +  ^'  —  2a6. 


186  ALGEBRA. 


Square  Roots  of  Compound  Expressions. 

208.  Since  the  square  of  a-{-b  is  a^-{-2ab-{-  J^,  the 
square  root  of  a^ -\- 2  ab -}- h^  is  a -\- b. 

It  is  required  to  find  a  method  of  extracting  the  root 
a  -f  ^  when  a^ -\- 2  ab -{- b^  is  given : 

of  the  root  is  obviously  the  square  root  of 
the  first  term,  a^,  in  the  expression. 

If  the  a2  is  subtracted  from  the 
given  expression,  the  remainder  is 
2ab  +  b^.  Therefore  the  second  term, 
6,  of  the  root  is  obtained  when  the 
first  term  of  this  remainder  is  divided 
by  2  a,  that  is,  by  double  the  part  of 
the  root  already  found.  Also,  since  2  ab  +  b^  =  {2  a  +  b)  b,  the  divisor 
is  completed  by  adding  to  the  trial-divisor  the  new  term  of  the  root. 

Ex.    Find  the  square  root  of  25  x^  —  20  x^f/  +  4  x*i/^, 

25x''  —  20xhj-{-Axy\5x--2^ 

2^7? 


Ex. 

The  first  term,  a,  of 

a?^2ab-\-b''\a^b 

a-^b 

2ab^b'- 

2ab^b^ 

10x  —  2xhj 


-^1       l/-{-4:X'^y^ 
—  20x^1/ -{-4:xy 


The  expression  is  arranged  according  to  the  ascending  powers  of  x. 

The  square  root  of  the  first  term  is  5  x,  and  5  x  is  placed  at  the 
right  of  the  given  expression,  for  the  first  term  of  the  root. 

The  second  term  of  the  root,  —2x^y,  is  obtained  by  dividing 
—  20  x^y  by  10  x,  and  this  new  term  of  the  root  is  also  annexed  to  the 
divisor,  10  x,  to  complete  the  divisor. 

209.  The  same  method  will  apply  to  longer  expressions, 
if  care  be  taken  to  obtain  the  trial-divisor  at  each  stage  of 
the  process,  by  doubling  the  part  of  the  root  already  found; 
and  to  obtain  the  complete  divisor  by  annexing  the  new  term 
of  the  root  to  the  trial-divisor. 


INVOLUTION    AND    EVOLUTION. 


187 


Ex.   Find  the  square  root  of 

16  a;«— 24  x«+25  x^— 20  x^+lO  a;^— 4  x+1 1 4  x^—Sx''-\-2x—l 


16  x' 
Sa^—Sx^ 


■24«»+25a;* 
-24ic«+  9x* 


8x^—6x^-\-2x 


lQx'-20x^-{-10a^ 
16x*-12x^+  4a;2 


Sx^—6x^-\-4.x—l 


-  80^3+  6a:2-4a;+l 

—  8:c3+  Gx2— 4x+l 


The  expression  is  arranged  according  to  the  descending  powers 
of  X. 

It  will  be  noticed  that  each  successive  trial-divisor  may  be  obtained 
by  taking  the  preceding  complete  divisor  with  its  last  term  doubled. 


Exercise  78. 


v'/ 


Extract  the  square  roots  of:     ^.r 

1.  a^  +  4a3+2a  — 4^  +  1.       "^ 

2.  x''  —  2x''y  +  ^xhf  —  2xif  +  y\ 

3.  4a«—12a«x +  5aV  +  6aV  +  aV. 

4.  9a;«-12rcy  +  16icy-24ay  +  4/  +  16x/. 

5.  4a«+16c«+16aV--32aV. 

6.  4x*  +  9— 30a;  — 20ic»+37£c2 

7.  16  ic^  — 16  aZ»ic2  -f  16  ^,2^2  _[.  4  ^2^2  _  8  a^,8  +  4  6*. 

8.  ic«  +  25x2  +  10x^  —  4a«  — 20 .r^+ 16  —  240!. 

9.  it;«  +  8xy  — 4a;«y  — 4j-/+8.^y  — 10a;y  +  /. 

10.  4  — 12a  — llaH5a'-4««  +  4a«  +  14a3.. 

11.  9a''-'Qab  +  Z0ac-\-Qad  +  V  —  10hc  —  2hd 

+  25c2+10ct^+c?2. 


188  ALGEBRA. 

12.  25 x^  —  31  a^y  +  34 xhf -^Ox'y-\-f  —  ^xif^lO x^. 

13.  m^  — 4m^  +  10m«  — 20m^  — 44??i^ 

+  35  m^  +  46  m^  -  40  m  +  25. 
7 

14.  cc'*  —  x^y  —  -  i^y  +  xy^  -f  ^Z''- 

15.  £c^  — 4a;3y+6a;y  — 6a:?/^+5/  — ^  +  4 

X  X 

a^      a?x  ,  43   „  „      3      ^  ,  x^ 
''•    9-T  +  48""  -r"+4• 
t//  C'C  »^  «A/  tX>  fcC- 

a«      2(1  ,   „       2J   ,   J''  ,  ,     ,      6a;2      a;   ,   1 


Square  Roots  of  Arithmetical  Numbers. 

210.  In  the  general  method  of  extracting  the  square 
root  of  a  number  ei2>^^sed  by  figures,  the  first  step  is  to 
mark  off  the  figures  in  yroi^fs. 

Since  1  =  1^,  100  =  lO'^,  10,000  =  100^,  and  so  on,  it  is  evident  that 
the  square  root  of  any  number  between  1  and  100  lies  between  1  and 
10 ;  the  square  root  of  any  number  between  100  and  10,000  lies  be- 
tween 10  and  100.  In  other  words,  the  square  root  of  any  integral 
number  expressed  by  one  or  two  figures  is  a  number  of  one  figure  ; 
the  square  root  of  any  integral  number  expressed  by  three  or  four 
figures  is  a  number  of  two  figures  ;  and  so  on. 

If,  therefore,  an  integral  number  be  divided  into  groups  of  two 
figures  each,  from  right  to  left,  the  number  of  figures  in  the  root  will 
be  equal  to  the  number  of  groups  of  figures.  The  last  group  to  the 
left  may  have  one  or  two  figures. 

Ex.     Find  the  square  root  of  3249. 

32  49  (57  jn  this  case,  a  in  the  typical  form  a^  +  2  a6  +  6^ 

^^  represents  5  tens,  or  50,  and  b  represents  7  ones. 

107)  7  49  The  25  subtracted  is  really  2500,  that  is,  a2,  and 

7  49  the  complete  divisor,  2  a  +  6,  is  2  x  50  +  7,  or  107. 


INVOLUTION    AND    EVOLUTION.  189 

211.  The  same  method  will  apply  to  numbers  of  more 
than  two  groups  by  considering  a  in  the  typical  form  to 
represent  at  each  step  the  part  of  the  root  already  found. 

It  must  be  observed  that  a  represents  so  many  tens  with  respect  to 
the  next  figure  of  the  root. 

Ex.   Find  the  square  root  of  5,322,249. 

5  32  22  49(2307 
4 


43J132 
129 
4607)32249 
32249 

212.  If  the  square  root  of  a  number  have  decimal  places, 
the  number  itself  will  have  twice  as  many. 

Thus,  if  0.21  be  the  square  root  of  some  number,  this  number  will 
be  (0.21)2  =  0.21  X  0.21  =  0.0441 ;  and  if  0.111  be  the  root,  the  num- 
ber will  be  (0.111)2  =  0.111  X  0.111  =  0.012321. 

Therefore,  the  number  of  decimal  plac^  in  every  square  decimal 
will  be  even,  and  the  number  of  decimal  places  in  the  root  will  be 
half  as  many  as  in  the  given  number  itself. 

Hence,  if  the  given  square  nimiber  contains  a  decimal,  we  divide  it 
into  groups  of  two  figures  each,  by  beginning  at  the  decimal  point 
and  marking  toward  the  left  for  the  integral  number,  and  toward 
the  right  for  the  decimal.  We  must  be  careful  to  have  the  last  group 
on  the  right  of  the  decimal  point  contain  two  figures,  annexing  a 
cipher  when  necessary. 

Ex.      Find  the  square  roots  of  41.2164  and  965.9664. 

41.21  64  (6.42  9  65.96  64)  31.08 

36  9 

I24J52T  6lj65 

496  61 

1282)2564  6208)49664 

2564  49664 

It  is  seen  from  the  grouping  that  the  root  of  the  first  example  will 
have  one  integral  and  two  decimal  places,  and  that  the  root  of  the 
second  example  will  have  two  integral  and  two  decimal  places. 


190  ALGEBRA. 

213.  If  a  number  contains  an  odd  number  of  decimal 
places,  or  if  any  number  gives  a  remainder  when  as  many- 
figures  in  the  root  have  been  obtained  as  the  given  number 
has  groups,  then  its  exact  square  root  cannot  be  found. 
We  may,  however,  approximate  to  its  exact  root  as  near  as 
we  please  by  annexing  ciphers  and  continuing  the  operation. 

Ex.   Find  the  square  roots  of  3  and  357.357. 


3.  (1.732 

3  57.35  70(18.903 

1 

1 

27)200  ■ 

28)257 

189 

224 

343)1100 

369)3335 

1029 

3321 

3462)7100 

37803) 147000 

6924 

113409 

Exercise  79. 
Extract  the  square  roots  of : 

1.  120,409;  4816.36;   1867.1041;  1435.6521;   64.128064. 

2.  16,803.9369;   4.54499761;  0.24373969;  0.5687573056. 

3.  0.9;  6.21;   0.43;  0.00852;   17;   129;   347.259. 

4.  14,295.387;   2.5;   2000;   0.3;   0.03;   111. 

5.  0.00111;   0.004;   0.005;   2)   b]   3.25;   8.6. 

6.  i;  if;  m;  Ml;  III;  IM- 

7        1.2.3.        1.  7.         fi..6.        1. 

Cube  Eoots  of  Compound  Expressions. 

214.  Since  the  cube  of  a  +  ft  is  a^  +  3  a^j  +  3  ah^  +  b^,  the 
cube  root  of  a^  +  3  a'^h  +  3  a^^  -j-  ^^  is  a  +  h. 

It  is  required  to  find  a  method  for  extracting  the  cube 
root  a-\-h  when  a^-{-3a%-\-3  a1/  +  h^  is  given  : 


INVOLUTION   AND    EVOLUTION.  191 

(1)   Find  the  cube  root  of  a^  +  3  a'b  -\-^ab''-\-  b\ 


8  a' 


+  3ab-{-b' 


3a'  +  Sab  +  F 


3a%-\-3ab^-}-b^ 
3a^-\-3ab^  +  b^ 


The  first  term  a  of  the  root  is  obviously  the  cube  root  of  the  first 
term  a^  of  the  given  expression. 

If  a^  is  subtracted,  the  remainder  is  S  a-b  +  S  ah^  i- b^  ;  therefore, 
the  second  term  &  of  the  root  is  obtained  by  dividing  the  first  term  of 
this  remainder  by  three  times  the  square  of  a. 

Also,  since  Sa^b-\-S ab^-\-  b^=  {Sa^+ Sab  +  b"-) 6,  the  complete 
divisor  is  obtained  by  adding  Sab-\-  b'^  to  the  trial  divisor  3  a^. 

(2)    Find  the  cube  root  of  8  a;^  +  36  x'l/  +  54  :f y^  +  27  if. 

Sx^+36x'i/+54.xi/-\-27i/  \2x-\-3y 


(6a;+3y)3y=        -\-lSxy+9i/ 


12a;24-18x?/+9?/^ 


36x'f/-{-54:Xir-\-27f 
36x'i/-\-54.xf-{-27f 


The  cube  root  of  the  first  term  is  2  x,  and  this  is  therefore  the  first 
term  of  the  root. 

The  second  term  of  the  root,  3  y,  is  obtained  by  dividing  30  x^y  by 
3(2x)2=:  12cc2,  which  corresponds  to  Sa^  in  the  typical  form,  and  is 
completed  by  annexing  to  12  x-  the  expression  {3(2x)  +  3?/}3?/  =18xy 
+  9  2/2,  which  corresponds  to  3  a6  +  6^,  in  the  typical  form. 

215.  The  same  method  may  be  applied  to  longer  expres- 
sions by  considering  a  in  the  typical  form  3  a' -\- 3  ab -\- b' 
to  represent  at  each  stage  of  the  process  the  part  of  the  root 
already  found. 

Thus,  if  the  part  of  the  root  already  found  be  x  +  y,  then  3  a^  of 
the  typical  form  will  be  represented  by  3  (x  +  yY ;  and  if  the  third 
term  of  the  root  be  +  z,  the  3  a&  +  6'^  will  be  represented  by  3  (x  +  y)z 
+  z2.  So  that  the  complete  divisor,  3  a^  +  3  a6  +  6^^  ^ijl  be  repre- 
sented by  3  (X  +  2/)^  +  3  (X  +  y)  2  +  z'^. 


192  ALGEBRA. 

Ex.     Find  the  cube  root  oi  x^  —  Sx^  +  Bx^—Sx  —  l. 
.  3  X*  x^ 


{Sx'^-x)i-x)=        -3x3+    a;2 


3x4-3x3+    x2 


-3x5  +  5x8 

-3x5        +3x*-    x3 


3  (x2  -  x)2=3  x4  —  6  x3  +  3  x2 
(3x2-3  x-1)  (-1)= -3x2+3x  +  l 


3x4-0:88  +3x  +  l 


3x4  +  6x3-3x-l 


-3x4  +  6x3-3x-l 


The  root  is  placed  above  the  given  expression  for  convenience  of 
arrangement. 

The  first  term  of  the  root,  x2,  is  obtained  by  taking  the  cube  root 
of  the  first  term  of  the  given  expression  ;  and  the  first  trial-divisor, 
3  x*,  is  obtained  by  taking  three  times  the  square  of  this  term  of  the 
root. 

The  first  complete  divisor  is  found  by  annexing  to  the  trial-divisor 
(3x2  — x)(— x),  which  expression  corresponds  to  (3  a +  6)  6  in  the 
typical  form. 

The  part  of  the  root  already  found  (a)  is  now  represented  by  x2  —  x ; 
therefore,  3  a'^  is  represented  by  3  (x2  —  x)2  =  3  x*  —  6  x^  +  3  x2,  the  sec- 
ond trial  divisor  ;  and  (3  a  +  &)  6  by  (3  x2  —  3  x  —  1)  (—  1) ;  therefore, 
in  the  second  complete  divisor,  3  a^  +  (3  a  +  6)  6  is  represented  by 
(3x4  -  6x3  +  3x2)  +  (-  3x2-  3x  -  1)  X  (-  1)  =  3x4  -  6x8+  3x  +  1. 

Exercise  80. 
Find  the  cube  roots  of : 

1.  x^-^Qx''y-{-12xf'Y^if     3.    a;3  +  12a;2  +  48.'r +  64. 

2.  a^  — 9a2  +  27a  — 27.  4.    x^—^ax^-\-ba^Tc^—^a^x—a\ 

5.  a;«  +  3x-*  +  6a^*  +  7a:^  +  6:c2_|_3^_^l_ 

6.  1  — 9a;  +  39a;2  — 99a:3+156a;*  — 144a;''  +  64a;«. 

7.  a«— ^a^  +  9a^  +  4a3  — 9a2  — 6a  — 1. 

8.  64a;«+192a;*  +  144x*-32a;3-36a;2+ 120^-1. 

9.  1— 3a:+6cc2— 10^3+12x*  — 12x^  +  10x«  — 6iz;^+3x«-x9. 


INVOLUTIOX    AND    EVOLUTIOX.  193 

10.  a^-\-9a'b  — 13d  a^b'-i- 729  ab'  — 729  b\ 

11.  c^— 12 bc^-\- 60 bh'  — 160 b'c^-\- 24.0 b'f^  — 192 b'c  -f  6U\ 

12.  8  a''  +  48  a'b  +  60  «^^2  _  go  a^b^  —  90  a^^^  + 108  ab'  —  27  b^ 

Cube  Roots  of  Arithmetical  Numbers. 

216.  In  extracting  the  cube  root  of  a  number  expressed 
by  figures,  the  first  step  is  to  mark  it  off  into  groups. 

Since  1  =  l^,  1000=  lO^,  1,000,000=  lOO^,  and  so  on,  it  follows 
that  the  cube  root  of  any  integral  number  between  1  and  1000,  that  is, 
of  any  integral  number  which  has  one,  two,  or  three  figures,  is  a  num- 
ber of  one  figure  ;  and  that  the  cube  root  of  any  integral  number 
between  1000  and  1,000,000,  that  is,  of  any  integral  number  which 
has  four,  five,  or  siz  figures,  is  a  number  of  two  figures  ;  and  so  on. 

If,  therefore,  an  integral  number  be  divided  into  groups  of  three 
figures  each,  from  right  to  left,  the  number  of  figures  in  the  root  will 
be  equal  to  the  number  of  groups.  The  last  group  to  the  left  may 
have  one,  two,  or  three  figures. 

217.  If  the  cube  root  of  a  number  contain  any  decimal 
figures,  the  number  itself  will  contain  three  times  as  many. 

Thus,  if  0.3  is  the  cube  root  of  a  number,  the  number  is 
0.3  X  0.3  X  0.3  =  0.027. 

Hence,  if  the  given  cube  number  have  decimal  places,  we  divide 
it  into  groups  of  three  figures  each,  by  beginning  at  the  decimal  point 
and  marking  toward  the  left  for  the  integral  number,  and  toward 
the  right  for  the  decimal.  We  must  be  careful  to  have  the  last  group 
on  the  right  of  the  decimal  point  contain  three  figures,  annexing 
ciphers  when  necessary. 

If  the  given  number  be  not  a  perfect  cube,  ciphers  may  be  annexed, 
and  a  value  of  the  root  may  be  found  as  near  to  the  true  value  as  we 
please. 

218.  It  is  to  be  observed  that  if  a  denotes  the  first  term 
of  the  root,  and  b  the  second  term,  the  first  complete  divisor 
is  3a^  +  3ab-{-l)', 

and  the  second  trial-divisor  is  3  (a  -J-  ^)^  that  is, 
3a^-^6ab-\-M\ 


194 


ALGEBRA. 


which  may  be  obtained  from  the  preceding  complete  divisor 
by  adding  to  it  its  second  term  and  twice  its  third  term: 

a  method  which  will  very  much  shorten  the  work  in  long 
arithmetical  examples. 

219.   Ex.    Extract  the  cube  root  of  5  to  five  places  of 
decimals. 

5.000(1.70997 

1 

3X102  =  300 


3  (10X7)  =  210 
7^=   49- 
^59 
259  J 
3X1700^  =  8670000 
3(1700X9)=     45900 
92=  81 


4  000 


3  913 


8715981  I 
45981 J 
3X1709^  =  8762043 


87  000  000 


78  443  829 


8  om  171  0 

7  885  838  7 


670  332  30 
613  343  01 


After  the  first  two  figures  of  the  root  are  found,  the  next  trial- 
divisor  is  obtained  by  bringing  down  the  sum  of  the  210  and  49 
obtained  in  completing  the  preceding  divisor ;  then  adding  the  three 
lines  connected  by  the  brace,  and  annexing  two  ciphers  to  the  result. 

The  last  two  figures  of  the  root  are  found  by  division.  The  rule 
in  such  cases  is,  that  two  less  than  the  number  of  figures  already 
obtained  may  be  found  without  error  by  division,  the  divisor  to  be 
employed  being  three  times  the  square  of  the  part  of  the  root  already 
found. 


involution  and  evolution.  195 

Exercise  81. 
Find  the  cube  roots  of : 

1.  274,625.               7.    1601.613.  13.  33,076.161. 

2.  110,592.               8.    1,259,712.  14.  102,503.232. 

3.  262,144.               9.    2.803221.  15.  820.025856. 

4.  884.736.      10.  7,077,888.  16.  8653.002877. 

5.  109,215,352.   11.  12.812904.  17.  1.371330631. 

6.  1,481,544.    12.  56.623104.  18.  20,910.518875. 

19.    91.398648466125.            20.  5.340104393239. 

21.    Find  to  four  figures  the  cube  roots  of  2.5;  0.2;  0.01; 
4;  0.4. 

220.  Since  the  fourth  power  is  the  square  of  the  square, 
and  the  sixth  power  the  square  of  the  cube ;  the  fourth  root 
is  the  square  root  of  the  square  root,  and  the  sixth  root  is  the 
cube  root  of  the  square  root.  In  like  manner,  the  eighth, 
ninth,  twelfth roots  may  be  found. 

Exercise  82. 
Find  the  fourth  roots  of: 

1.  81  a^  —  540  a%  +  1350  a-^^  — 1500  ah^  +  625  h\ 

2.  l-4a:  +  10a;2-16x«+19x*-16a-«  +  10a;<^-4^^  +  a;«. 

Find  the  sixth  roots  of : 

3.  64  — 192cc  +  240x2  — 160a;3  +  60a;^  — 12aj«  +  a^'. 

4.  729 a:«- 1458 :tH  1215:c-*- 540a:«-f  135a;2- 18a^  + 1. 

Find  the  eighth  root  of : 

5.  l-82/-f28y2-562/^  +  70y-56/  +  282/^-8/  +  .y». 


CHAPTER  XIV. 
Quadratic  Equations. 

221.  An  equation  which  contains  the  square  of  the 
unknown  quantity,  but  no  higher  power,  is  called  a 
quadratic  equation. 

222.  If  the  equation  contains  the  square  only,  it  is  called 
a  pure  quadratic ;  but  if  it  contains  the  first  power  also,  it  is 
called  an  affected  quadratic. 

Pure  Quadratic  Equations. 

(1)    Solve  the  equation  5 ic^  —  48  =  2a;^. 

6x^  —  48  =  2  X'^  It  will  be  observed  that  there  are  two  roots  of 

3  ^2  __  ^g       equal  value  but  of  opposite  signs  ;  and  there  are 

2      ^n       only  two,  for  if  the  square  root  of  the  equation, 

iC      —  Id  c  ^r,  ...  .  1       ^        .,  Ill 


•.a;  =  ±4 


x2  =  16,  vsrere  written  ±  x  =  ±4,  there  would  be 
only  two  values  of  x ;    since   the  equation  —  x 


=  +  4  gives  X  =  —  4,  and  the  equation  —  x  =  —  4  gives  x  =  4. 

Hence,  to  solve  a  pure  quadratic. 

Collect  the  unknown  numbers  on  one  side,  and  the  known 
numbers  on  the  other;  divide  by  the  coefficient  of  the 
unknown  7iumber;  and  extract  the  sqiiare  root  of  each  side 
of  the  resulting  equation. 

(2)    Solve  the  equation  3  ic^  — 15  =  0. 

oar      15  =  0  j^  ^^^  -^^  observed  that  the  square  root  of  5 

3if^  =  15  cannot  be  found  exactly,  but  an  approximate 

cc^  =  5  value  of  it  to  any  assigned  degree  of  accuracy 

.*.  ic  =  =h  V5  °^^y  ^^  found. 


QUADRATIC    EQUATIONS.  197 

\.   A  root  which  is  indicated,  but  which  can  be  found 
only  approximately,  is  called  a  Surd. 

Ex.    Solve  the  equation  3  a;^  + 15  =  0. 

oar-t-lo  =  U  j^.  ^-^Yi  ^^g  observed  that  the  square  root 

3x^  =  —  15  of  —  5  cannot  be  found  even  approximately; 

x^  =  —  5  for  the  square  of  any  number,  positive  or 

.  • .  £C  =  =b  V 5  negative,  is  positive. 

224.    A  root  which  is  indicated,  but  which  cannot  be 
found  exactly  or  approximately,  is  imaginary.     §  206. 

Exercise  83. 
Solve : 

1.  x'-3  =  4:6.  6.    5x^-9  =  2x^  +  24.. 

2.  2(a;2— 1)— 3(ct;2+l)+14=0.  7.    (x-^2y  =  4x-{-5. 

x^      x^—l(i_^      m+x' 
®*    5  15     ~'  25    ' 

3x^—27      90  +  4x2^ 

10.    ^x-\--  =  —=-' 
X        7 

4x^4-5      2a;^  — 5  _  7x^  —  25 
^^*        10  15      ~       20      * 

10^2  +  17      12a:2_|_2      5a^  —  4: 


3. 

x^-5      2x^+1       1 

3       '        6       ~2 

4. 

3       ,       3     _, 

3         17 

b. 

4:x^      6a;2~3 

12. 


13. 


18  11a;-- 8  9 

14ic2+16       2a;2  +  8       2x^ 


21  8x^  —  11        3 

14.    x^ -\- bx -{- a  =  bx  (1  —  bx). 
15.    mx^ -\-n  =  q.  IQ.    x^  —  ax-\-b  =  ax(x  —  1). 


19S  ALGEBRA. 

Affected  Quadratic  Equations.  ■ 

225.  Since  (ax  ±  by  =  a?Qi?  d=  2  ahx  +  ^^  it  is  evident  that 
the  expression  ii?x^±2ahx  lacks  only  the  third  term,  h"^,  of 
being  a  complete  square. 

It  will  be  seen  that  this  third  term  is  the  square  of  the 
quotient  obtained  from  dividing  the  second  term  by  twice  the 
square  root  of  the  first  term. 

226.  Every  affected  quadratic  may  be  made  to  assume 
the  form  of  A^  ±  2  abx  =  e. 

The  first  step  in  the  solution  of  such  an  equation  is  to 
complete  the  square;  that  is,  to  add  to  each  side  the  square 
of  the  quotient  obtained  from  dividiiig  the  second '  term  by 
twice  the  square  root  of  the  first  term. 

The  second  step  is  to  extract  the  square  root  of  each  side 
of  the  resulting  equation. 

The  third  and  last  step  is  to  reduce  the  resulting  simple 
equation. 

(1)    Solve  the  equation  16  a?^  +  5  a?  —  3  =  7  cc^  —  a;  -}-  45. 

16x2  +  5x  -  3  =  7  x2  -  X  +  45. 

Simplify,  9x2+6x  =  48. 

Complete  the  square,  9cc2  +  6x  +  1  =  49. 
Extract  the  root,  3ic'+  1  =  ±  7. 

Reduce,  3x=  —  l  +  7or  —  1  —  7, 

3  X  =  6  or  —  8. 
.-.  X  =  2  or  —  2f . 

Verify  by  substituting  2  for  x  in  the  equation 

16x2+5x-3  =  7x2-x  +  46, 
16  (2)2  +  5  (2)  -  3  =  7  (2)2  -  (2)  +  46, 
«4  +  10  -  3  =  28  -  2  +  45, 
71":  71. 


QUADRATIC    EQUATIONS.  199 

T«rify  by  substituting  —  2|-  for  x  in  the  equation 

16x2 +  5x- 3=  7x2-x  +  45, 

16(-f)2+5(-f)-3  =  7(-|)2-(-|)+45., 

io^  -  -V-  -  3  =  A41  +  I  +  45, 

1024  -  120  -  27  =  448  +  24  +  406, 

877  =  877. 


(2)    Solve  the  equation  3x^  —  4:X  =  32. 

Since  the  exact  root  of  3,  the  coefficient  of  x^,  cannot  be  found,  it 
is  necessary  to  multiply  or  divide  each  term  of  the  equation  by  3  to 
make  the  coefficient  of  x^  a  square  number. 

Multiply  by  3,  9  x2  -  12  x  =  96. 

Complete  the  square,  9  x2  —  12  x  +  4  =  100. 
Extract  the  root,  3  x  —  2  =  ±  10. 

Reduce,  8x  =  2  +  10  or  2  -  10  ; 

3x=:12or— 8. 
.-.  X  =  4  or  -  2f . 


Or,  divide  by  3,  ^^~'J~  J' 

^        ,        ,  o      4x  ,  4      32  ,  4      100 

Complete  the  square,        x^ —+-  =  -—  +  -=     -  • 

2       ,  10 
Extract  the  root,  x  —  -  =  ±  -r  • 


.'.  X 


3  3 

2  ±10 


3 
=  4  or  -  2f . 


Verify  by  substituting  4  for  x  in  the  original  equation, 

48  -  16  =  32, 
32  =  32. 

Verify  by  substituting  —  2f  for  x  in  the  original  equation, 

21i-(-10f)  =  32, 
32  =  32. 


200  ALGEBRA. 

(3)    Solve  the  equation  —3x^-]-5x  =  —  2.  k 

Since  the  even  root  of  a  negative  number  is  impossible,  it  is  neces- 
sary to  change  the  sign  of  each  term.     The  resulting  equation  is, 


3x2 

— 

5x  =  2. 

Multiply  by  3, 

9x2-15x=6. 

Complete  the  square. 

9: 

C2- 

-.r      .25      49 
-16^  +  7=4- 

Extract  the  root, 

5    ^: 

«^-2=±2- 

Reduce, 

5±7 
3x=     2     . 

3x  =  6  or  —  1. 
.-.  X  =  2  or  —  i. 

Or,  divide  by  3, 

,      5x      2 

Complete  the  square, 

X^ 

,      5x25      49 
3  "^36      36* 

Extract  the  root, 

5       ,   7 

5±7 
.•.x=      g     , 

=  2  or  -  i. 

If  the  equation  3  x2  —  5  x  =  2  be  multiplied  by  four  times  the  coeffi- 
cient of  x2,  fractions  will  be  avoided  : 

36x2  — 60x  =  24. 
Complete  the  square,  36  x2  —  60  x  +  25  =  49. 
Extract  the  root,  6  x  —  5  =  ±7, 

6x  =  5  ±  7, 
6x=  12  or  —  2. 
.-.  X  =  2  or  —  -J-. 

It  will  be  observed  that  the  number  added  to  complete  the  square 
by  this  last  method  is  the  square  of  the  coefficient  of  x  in  the  original 
equation  3  x2  —  5  x  =  2. 


QUADKATIC    EQUATIONS.  201 


(4)  Solve  the  equation ;- ^  =  2. 

Simplify  (as  in  simple  equations), 

4x2  — 23x=  -30. 
Multiply  by  four  times  the  coefficient  of  x^,  and  add  to  each 
side  the  square  of  the  coefficient  of  x, 

64  x2  —  (  )+  (23)2  =  529  -  480  =  49. 
Extract  the  root,  8  x  —  23  ==  ±  7. 

Reduce,  8x  =  23±7; 

8x=30orl6. 
.-.  X  =  3f  or  2. 

If  a  trinomial  is  a  perfect  square,  its  square  root  is  found  by  taking 
the  square  roots  of  the  first  and  third  terms  and  connecting  them  by 
the  sign  of  the  middle  term.  It  is  not  necessary,  therefore,  in  com- 
pleting the  square,  to  write  the  middle  term,  but  its  place  may  be 
indicated  as  in  this  example. 

(5)  Solve  the  equation  72a;2  — 30a^  =  — 7. 

Since  72  =  2=^  X  3^,  if  the  equation  be  multiplied  by  2  the  coeffi- 
cient of  x2  in  the  resulting  equation,  144  x2  —  60  x  =  —  14,  will  be  a 
square  number,  and  the  term  required  to  complete  the  square  will  be 
(11)2  =  (5)2  =  2^.  Hence,  if  the  original  equation  be  multiplied  by 
4  X  2,  the  coefficient  of  x^  in  the  result  will  be  a  square  number,  and 
fractions  will  be  avoided  in  the  work. 

Multiply  the  given  equation  by  8, 

676x2-240x=  -56. 

Complete  the  square,  676x2  —  ( )  +  25  =  —  31^ 

Extract  the  root,  24  x  —  5  =  ±  V— 31. 

Reduce,  24x  =  5±V-31. 


.-.  X  =  ^\  (5  ±  V-  31). 

Note.     In  solving  the  following  equations,  care  must  be  taken  to 
select  the  method  best  adapted  to  the  example  under  consideration. 


Solve:  Exercise  84. 

1.  x^-\-4:X  =  12.         4.   a;2  — 7x  =  8.         ^7.  a;^  — cc  =  6. 

2.  x'^  —  6x  =  W.         5.    3x2  — 4ic  =  7.     ,.  g.  5x^  —  Sx  =  2. 

3.  a'2-12a-f-6=i.Z^6.    12a^2^.T-l=0^  9.  2x^-27x=U. 


202  ALGEBRA. 


/    ,„      2      2a;  ,    1        _  ,„     x  +  1      2a!  — 1 

^10.   .^--  +  -=0.  13.    _^  =  -^. 

11.   ^^-|  =  2(x  +  2).  14.       ^  ^  +  ^ 


2       3      -^-^  '  "^  *^-    ^  +  1      2(£c  +  4)  18 

3a;   ,     4       13  _         2  3.2 

12.    -r  +  7;-  =  Tr'  16. 


4     '  3x       6  ic  — 1      ic  — 2      X  — 4 

16.    5:r(;z;-3)~2(x2_g)^(^_|.3)(^_|.4). 

3cc  5_    ^x"  23 

^'^'    2(^  +  1)       8~a;2  — 1      4(x  — 1)* 

18.  (x  —  2)(x-4.)-2{x-l)(x-3)  =  0. 

19.  ^(^_4)-|(a3-2)  =  ^(2x  +  3). 

20.  \{Zx'-x-b)-\(x'-l)  =  2{x~2y. 
o  o 

2x        3^  —  50   _12a;  +  7Q 
^^*    15"^3(10  +  a^)"~      190 

X      _  15  — 7a;  _  14xj-9._  x'^  —  S 

22.  ^^171-8(1 -X)'  ^^-    ^       8a^-3~:r  +  l' 

2a;  — 1   ,   1      2a;  — 3        „^     .        a;  +  5        cc  — G 

23.  -^Z:r  +  6  =  ^^'       ''•    ^-2^+1^^^- 

a;  +  2       4-a;_7_  ^_    .   Izi^-oa 

24.  ^31-^^-3  2^-    7-X+     cc      -^^^• 

2a;  +  3  7  — a;         7  — 3a; 

28.  ^  — 


2(2a;  — 1)       2(a;  +  l)       4  —  3 


x 


12a;«-lla;^  +  10^-78_         _1 
^^-  8a;2-7a;  +  6"        ~^^^      2* 

6  18    ^     7  8 

X  —  1      a;  +  5      a;4-l      x  —  b 


QUADRATIC    EQUATIONS.  208 

327.   Literal  quadratic  equations  are  solved  as  follows  : 

(1)    Solve  the  equation  ax^  -\'bx  =  c. 

Multiply  the  equation  by  4  a  and  add  the  square  of  6, 

4a2x2  +  (  )  +  62  =  ^ac  +  b^. 
Extract  the  root,  2ax-\-b  =  ±  V4ac  +  6^. 

Keduce,  2ax=  —b±  V4  ac  +  &2. 

-  6  ±  V4  ac  +  62 


2a 
(2)    Solve  the  equation  ac?a;  —  acx^  =  bcx  —  bd. 

Transpose  bcx  and  change  the  signs, 

acx2  +  bcx  —  adx  =  bd. 

Express  the  left  member  in  two  terms, 

acx^  +  {be  —  ad)x  =  bd. 

Multiply  by  4  ac, 

4  a2c2x2  +  4  ac  (6c  —  ad)x  =  i  abed. 

Complete  the  square, 

4  a2c2x2  +  (  )  +  (6c  -  ad)2  =  62c2  +  2  a6cc«  +  a2d2. 

Extract  the  root,         2  acx  +  (6c  —  ad)  =  ±  {be  +  ad). 

Reduce,  2aex  =  —  (6c— ad)±(6c+ad), 

=  2  ad  or  —  2  6c. 

d  6 

.-.  X  =  -  or 

c  a 


(3)    Solve  the  equation  px^  ^  px -\- qx^ -\- qx 
Express  the  left  number  in  two  terms, 

(p  +  g)x2-(p-g)x  =  ^- 

Multiply  by  four  times  the  coefficient  of  x2, 

4  (p  +  g)2x2  —  4  (jp2  —  q2^x  =  4pq. 
Complete  the  square, 

4(p  +  7)2x2-(  )-\.{p-qy2  =  pi+2pq4-  ^  , 
Extract  the  root,  2  {p  -\-  q)  x  —  {p  —  q)  =  ±  {p  +  q). 
Reduce,  2  (p  + 5)x=  (p  — g)  ±  (p  +  g), 

=  2  p  or  —  2  g. 

p  q 

.:  X  —      ,      or -. — • 

P  +  g  p  +  g 

Note.  The  left-hand  member  of  the  equation  when  simplified 
must  be  expressed  in  two  terms,  simple  or  compound,  one  term  con- 
taining x2,  and  the  other  term  containing  x. 


204  ALGEBRA. 

Exercise  85. 
Solve : 

1.  x'^-{-2ax  =  a\  14.    x^-{-ax-=a-{-x. 

2.  x^  =  4:ax-\-7a\  15.    x^ -{- ax  =  bx -{- ab. 

„7m^^  X   ,   a      X   ,    b 

3.  x^=— Smx.  16.    -  +  -  =  -+  — 

4  a      X       0      X 

5nx      Sn^      .  ^^1,1  1    . 

4.  x^ ^=0.  17.    -4 


2  2  X      x-\-b      a      a-\-b 


b^ 


a  ,  5x       X- 


^'    (^  +  «)'       {x  —  af  ^^*    3~^4        3^      ^' 

9172  ^^  -.«^  +  ^  ,iC  — 3 

6.  cx=^ax^-\-bx^ 7—'  19.    -  =  c»H — -• 

a  +  o  a?  —  3  x-Y?> 

7.  -— +  -  = 20.    mx^—l  =  ^ ^• 

b^  (i^  C  7)171 

8.  {a'^-\-l)x=^ax^-{-a.  21.    (aa- —  ^>)  (te  —  a)  =  c^. 

a       ,       b  2  c  ax-\-b      7nx-\-7i 

9. ^ -== •  22.    r— J— = — X — 

X  —  a      X  —  0       X  —  c  ox-f-a      7ix-\-7n 

1  1,1,1  7n       ,       m 

10.  ——-—-  =  -  +  -  +  -  23.  — r-H =c. 

a-\-b-\-x       a       b      x  7n-\-x       in  —  x 

1  1  3  +  ic2  (a— l)V+2(3a— l)ic     , 

a  —  a;       a  +  x       a''  —  x^  4a — 1 

^2  +  ^  a-\-b^ 


13. 


2x  —  a-\-2b  (7n-{-7iy        ^  ^ 

\      a  —  b_14:a^-5ab-10b^      (2a  —  3b)x 
^^'    "^"^    ab^    ~  18^262  "T"        2ab 


QUADRATIC    EQUATIONS.  206 


28.    abx^-\ = ' — ^ 

c  cr  c 


29. 


i2  — 4a 


X 


Sm  —  2a      4a  —  6m      2 


30.  6x-\-^ — ■ — ^=5(a  — ^>)+-— — 

X  ^         ^        6x 

31.  f (a;2_|_^2^^^S)^i3,^20a  +  46). 

32.  x^  —  (b  —  a)c  =  ax  —  bx-\-cx. 

33.  x^  —  2  mx  =  (n  — ^;  +  m)  (n  — j^  —  ^)- 

34.  x^  —  (m  -\- n)  x  =  i  (2:f -\- q -\-  w  +  ^^)  (i?  +  ?  —  ^  —  ^). 

35.  m7w;^  —  (^  +  ^0  (^?i  +  1)  x  -f-  (_»i  +  ^)^  =  0. 

2  b  —  X  —  2a      4^  —  7a x  —  4a 

bx  ax  —  bx      ab  —  b^ 

37.    2x\a^-b'')-(Sa''-\-J)')(x-\)  =  {U^^a')(x^V), 

a  —  2b  —  x        5b  — X        2a  —  x  —  19b_ 
a^  —  4:b^         ax-\-2bx  2bx  —  ax      ~ 

x-\-lSa-\-3b      ^_a  —  2b 
5a  —  3b  —  X  x-\-2b 

x-^3b 3^ a-^3b 

Sa:'-12ab      9b'-4.a'      (2a-\-3b)  (x-3b)~~ 

41.  nx^-\-px — px^  —  vix-\-7n  —  n  =  0. 

42.  (a-\-b-\-c)x^  —  (2a-\-b-\-G)x  +  a  =  0. 

43.  (ax  —  b)  (c  —  d)  =  (a  —  b)  (ex  —  d)  x. 

2a;  +  l  _  1  A  __  2\  _  3a;  +  l. 
b  x\b       a  J  a 

1  ■  1  _       Q^  2bx-{-b 

'    2x^-\-x  —  l^  2x''  —  3x-\-l~  2bx  —  b       ax^  —  a 


44. 


206  ALGEBRA. 

228.  An  affected  quadratic  may  be  reduced  to  the  form 
x^  •\-px  -f-  2'  =  0,  in  which  p  and  q  represent  any  numbers, 
positive  or  negative,  integral  or  fractional. 

Ex.    Solve:  x^-\-px-{-q  =  0. 


.\x=:^-^d^-\lp^-4.q. 

By  this  formula,  the  values  of  x  in  an  equation  of  the 
form  x^-\-px-{-  q  =  0,  may  be  written  at  once.  Thus,  take 
the  equation 

Divide  by  3,       a:2  —  |  x  +  |  =  0. 
Here,  'P—~^i  ^-nd  q  —  ^. 


—   5  4-   1 

=  1  or  f . 

229.  A  quadratic  which  has  been  reduced  to  its  simplest 
form,  and  has  all  its  terms  written  on  one  side,  may  often 
have  that  side  resolved  by  inspection  into  factors. 

In  this  case,  the  roots  are  seen  at  once  without  com- 
pleting the  square. 

(1)    Solve  £c2  + 7. T  — 60  =  0. 

Since  x'-  +  7  x  —  60  =  (x  +  12)  (x  -  5), 

the  equation  x-  +  7  x  —  60  =  0 

may  be  written       (x  +  12)  (x  —  5)  =  0. 

It  will  be  observed  that  if  either  of  the  factors  x  +  12  or  x  —  5  is  0, 
the  product  of  the  two  factors  is  0,  and  the  equation  is  satisfied. 

Hence,        x  +  12  =  0  and  x  —  5  =  0. 
.-.  X  =  —  12,  and  x  =  5. 


QUADRATIC    EQUATIONS.  207 

(2)    Solveir2_|_7^^Q 

The  equation  x2  ^-  7  ^  =  0 

becomes  x  (x  +  7)  =  0, 

and  is  satisfied  if  x  =  0,  or  if  x  +  7  =  0. 

.'.  the  roots  are  0  and  —  7. 

It  will  be  observed  that  this  method  is  easily  applied  to  an  equation 
all  the  terms  of  which  contain  x. 


(3)  Solve  2x^  —  x^  —  6x  =  0. 

The  equation  2x^  —  x^  —  6x  =  0 

becomes  x  (2  x^  —  x  —  6)  =  0, 

and  is  satisfied  if  x  =  0,  or  if  2  x^  —  x  —  6  =  0. 

By  solving  2  x^  —  x  —  6  =  0  the  two  roots  2  and  —  f  are  found. 

.•.  the  equation  has  three  roots,  0,  2,  —  ^. 

(4)  Solvea;«  +  a;2  — 4a;  — 4  =  0. 

The  equation  x^+x^  —  4x  —  4  =  0 

becomes  x^  (x  +  1)  —  4  (x  +  1)  =  0, 

(x2-4)(x+ 1)  =  0. 

.*.  the  roots  of  the  equation  are  —  1,  2,  —  2. 


(5)    Solvea;3-2a;2  — lla;4-12  =  0. 

x8-2x2-llx+12 

Smce =  x2  —  X  —  12, 

X—  1 

the  equation  x8  —  2  x2  -  11  x  +  12  =  0 

may  be  written  (x  —  1)  (x^  —  x  —  12)  =  0. 

The  three  roots  are  found  to  be  1,  —  3,  4. 

An  equation  which  cannot  be  resolved  into  factors  by  inspection 
may  sometimes  be  solved  by  guessing  at  a  root  and  reducing  by  divi- 
sion. In  this  case,  if  a  denote  the  root,  the  given  equation  (all  the 
terms  of  the  equation  being  written  on  one  side)  may  b«  divided  by 
X  — a. 


208  ALGEBRA. 

Exercise  86. 
Find  the  roots  of  : 

1.  (x-\-l)(x—2)(x'+x-2)  =  0.  7.  x^  —  x^  —  x  +  l  =  0. 

2.  (x^—3x-^2)(x^—x  —  12)=0.  8.  Sa?^  — 1  =  0. 

3.  (x-\-l)(x-2)(x-}-3)=  —  6.  9.  8cc«  +  l  =  0. 

4.  2x^-{-4.x^—70x  =  0.  10.  a;«  — 1  =  0. 

5.  (x'  —  x  —  6)(x^  —  x  —  20)=0.  11.  cc(£c— a)(a;2-J2^)=0. 

6.  x(x+l)(x+2)=(a+2)(a-^l)a.      12.  7i(cc"^+l)+£c+l=0. 

230.    If  r  and  r'  represent  two  values  of  x,  then 

ic  —  r  =  0, 
and  x  —  7-'  =  0. 

.'.  (x  —  r)  (x  —  r')  =  0. 

This  is  a  quadratic  equation,  as  may  be  seen  by  performing  the 
indicated  multiplication. 

Now,  r  and  r'  are  roots  of  this  equation  ;  for,  if  either  r  or  r'  be 
written  for  x,  one  of  the  factors,  x  —  r,  x  —  r',  is  equal  to  0,  and  the 
equation  is  satisfied.  Also  r  and  r'  are  the  only  roots,  for  no  value 
of  X,  except  r  and  r',  can  make  either  of  these  factors  equal  to  0. 

Since  r  and  r'  may  represent  the  values  of  x  in  any  quadratic 
equation,  it  follows  that  every  quadratic  equation  has  two  roots,  and 
only  two. 

Again,  if  r,  r',  r'',  represent  three  values  of  x, 
then,  (x  —  r){x  —  r')  (x  —  r")  =  0. 

This  is  a  cubic  equation,  as  may  be  seen  by  performing  the  indi- 
cated multiplication.  Hence,  it  may  be  inferred  that  a  cubic  equation 
has  three  roots,  and  only  three;  and  so,  for  any  equation,  that  the 
number  of  roots  is  equal  to  the  degree  of  the  equation. 

It  may  also  be  inferred  that  if  r  be  a  root  of  an  equation,  x  —  r 
will  be  a  factor  cf  the  equation  when  the  equation  is  written  with  all 
its  terms  on  one  side. 


QUADRATIC    EQUATIONS.  209 

If  r  and  r'  represent  the  roots  of  the  general  quadratic  equation, 

x^  +  px+  q  =  0. 

This  equation  may  be  written       (x  —  r)  (x  —  r')  =  0, 
or,  x2  —  (r  +  r')  x  +  rr"  =  0. 

A  form  which  shows  that 
the  sum  of  the  roots  —  ~  Pt 

and  the  product  of  the  roots  =  q. 

231.  It  will  be  seen  from  §  230  that  an  equation  may  be 
formed  if  its  roots  are  known. 

If  the  roots  of  an  equation  be  —  1  and  i, 
the  equation  will  be  (x  4-  1)  (x  —  i)  =  0, 

or,  by  multiplying  by  4,  4  x2  +  3  x  —  1  =  0. 

If  the  roots  of  an  equation  be  0,  1,5, 
the  equation  will  be  (x  —  0)  (x  —  1)  (x  —  5)  =  0  ; 

that  is,  X  (X  —  1)  (X  -  5)  =  0, 

or,  x^  —  0  x-^  +  5  X  =  0. 

If  z  occur  in  every  term,  the  equation  will  be  satisfied  by  putting 
X  =  0,  and  may  be  reduced  to  an  equation  of  the  next  lower  degree 
by  dividing  every  term  by  x. 


By  considering  the  roots  of  x^-\-2yx-\-q  =  0, 
namely,  r  =  —l^-\--y/p^  —  4:q, 

and  r'  =  —  ^  —  -y/p^  —  4:q, 

it  will  be  seen  that  the  character  of  the  roots  of  an  equation 
may  be  determined  without  solving  it : 


I.  As  the  two  roots  have  the  same  expression,  y/jr  —  4  y, 
both  roots  will  be  real,  or  both  will  be  imaginary. 

If  both  be  real,  both  will  be  rational  or  both  surds,  accord- 
ing as  jo^ —  4  2'  is  or  is  not  a  perfect  square. 


210  ALGEBRA. 

II.  When  ^^  is  greater  than  4  q,  the  two  roots  will  be 
real,  for  then  the  expression  p^  —  4^'  is  positive,  and  there- 
fore V^^  —  4  2'  can  be  found  exactly  or  approximately. 

Since  also  its  value  in  one  root  is  to  be  added  to  —  ^^j 

and  in  the  other  to  be  subtracted  from  —  -t  the  two  roots 
will  be  different  in  value. 

III.  When  p^  is  equal  to  4  q.  the  roots  will  be  equal  in 
value. 

IV.  When  p'^  is  less  than  4  q,  the  roots  will  be  imaginary, 
for  then  the  expression  p>^  —  ^q  will  be  negative,  and  there- 
fore Vj?^  —  4  2'  represents  the  even  root  of  a  negative  number, 
and  is  imaginary. 

V.  If  q  (=  r  X  r')  be  positive,  the  roots,  if  real,  will  have 
the  same  sign,  but  opposite  to  that  of  ^  (since  r-\-r'^ — p). 

But  if  q  be  negative,  the  roots  will  have  opposite  signs. 

233.  Determine  by  inspection  the  character  of  the  roots  of: 
(1)    x2-5a^  +  6  =  0. 

In  this  equation  p  is  —  5,  and  q  is  6. 


.-.  Vp2  —  4 g  =  V25-24  =  1. 
.-.  the  roots  will  be  rational^  and  hath  positive. 

(2)    x'-]-3x-\-l  =  (). 

In  this  equation  p  is  3,  and  g-  is  1. 


...  Vp-^-4g  =  V9  — 4  =  Vs. 
.-.  the  roots  will  be  surds,  and  &o^/i  negative. 

(3)    a^2_p3a;  +  4  =  0. 

In  this  equation  p  is  3,  and  q  is  4. 

.-.  ^/p^  —  4:q  =  Vq  -  1(5  =  V^, 
.-.  the  roots  will  be  imaginary. 


quadratic  equations.  211 

Exercise  87. 
Form  the  equations  whose  roots  are : 

1.  2,1.  5.  -5,-i.  9.  0, --J,  1,-1. 

2.  7,-3.  6.  — ^,  f.  10.  a  —  2b,3a-i-2b. 

3.  i,i.  7.  3,  — 3,  f,  — f.  11.  2a  —  b,b  —  Sa. 

4.  f,  —  f.  8.  0,1,2,3.  12.  a(a+'l),l  —  a. 

Determine  by  inspection  the  character  of  the  roots  of : 

13.  x^  —  7x  +  12=0.  17.  x'-\-4.x-\-l=0. 

14.  £c2— 7x  — 30  =  0.  18.  a;2  — 2a;  +  9=:0. 

15.  x^-j-4.x  —  5  =  0.  19.  3a;2  — 4a:  — 4  =  0. 

16.  5a;2  +  8  =  0.  20.  a:2  +  4a;  +  4  =  0. 

234.   It  is  often  useful  to  determine  the  maximum  or 
minimum  value  of  a  given  quadratic  expression. 

(1)   Eind  the  maximum  or  minimum  value  of  1  +^  —  ^^• 

Let  \  +  x  —  x^=m; 

then,  x2  —  X  =  1  —  m, 

and  4x2  — (  )  +  1  =  5  — 4m, 


2a; -^  =  ±  V5  —  4 ! 


.-.  .'c  =  i±iV5-4m. 

Now,  for  all  possible  values  of  x,  5  —  4  m  cannot  be  negative ;  that 
is,  m  cannot  be  greater  than  | ;  and  for  this  value  x\s,\.  Therefore, 
f  is  the  maximum  value  of  the  given  expression. 

(2)    Find  the  maximum  or  minimum  value  of  £c^  +  3a:  +  4. 

Let  x2  +  3x-f4=wi; 

then,  x2  +  3  X  =  m  —  4, 

and  4x2+(  )+9=4m-7, 

2x  +  3=  ±  V4  m  -  7. 


X  =  —  I  ±  ■}V4  m  —  7. 

For  all  possible  values  of  x,  4  ?n  —  7  cannot  be  negative ;  that  is, 
m  cannot  be  less  than  | ;  and  for  this  value  x  =  —  f .  Therefore,  |  is 
the  minimum  value  of  the  given  expression. 


212  ALGEBRA. 

Exercise  88. 

Find  the  maximum  or  minimum  value  (and  determine 
which)  of: 

1.  4  +  6x  — a^l       4.    (a  —  x)(x  —  b).         7.    x^  — 2x  +  9. 

2.  (^  +  ^)'.     *         5.        ^     .  8.    ^^^ 

X  l-\-x^  '    (x-\-a)  (x — b) 

3.  ^^ii.  6.    ^2  + 8^^  +  20.  9.        "" 


a-\-x^ 


10.  Divide  a  line  20  in.  long  into  two  parts  so  that  the 

sum  of  the  squares  on  these  two  parts  may  be  the 
least  possible. 

11.  Divide  a  line  20  in.  long  into  two  parts  so  that  the 

rectangle  contained  by  the  parts  may  be  the  greatest 
possible. 

12.  Find  the  fraction  which  has  the  greatest  excess  over 

its  square. 

235.   Two  other  cases  of  the  solution  of  equations  bi/ 
comjpleting  the  square  should  be  noticed. 

I.  When  any  two  powers  of  x  are  involved,  one  of  which 
is  the  square  of  the  other. 

II.  When  the  addition  of  a  number  to  an  equation  of  the 
fourth  degree  will  make  both  sides  complete  squares. 

(1)    Solve  8cc«  +  63a:2  =  8. 

In  this  equation  the  exponent  6  is  the  double  of  3,  hence  x^  is 
the  square  of  x^. 

8jc6  +  63x3  =  8, 
266x6+ (  )+  (63)2=4226, 
16x8 +  63  =±65, 

16x8=2,  or -128, 
ic8  =  -J-,  or  -  8. 
By  taking  cube  root,  a;  =  f ,  or  —  2. 


QUADRATIC    EQUATIONS. 


213 


The  other  roots  of  the  equation  are  found  by  finding  the  remaining 
roots  of  the  equations,  x^  =  ^,  and  x^=  —  8. 


Since,  a;3  =  i,       .-.8x3-1=0 

Now,  by  §  230, 

8  a;3  _  1  =  (2  X  -  1 )  (4  x2  +  2  X  +  1 ) 

.-.  (2x-l)(4x2+2x+  iy=0 

and  is  satisfied  if4x2+2x+l  =  0 

as  well  as  if  2x— 1=0. 

-  The  solution  of4x2  4-2x+l=0 

gives    X  =  i  (-  1  ±  V^). 


Since,  x^  =  —  8,      .-.  x^  +  8=0 
Now,  by  §  230, 

x3  +  8  =  (X  +  2)  (x2  -  2  X  +  4) 

.-.  (x  +  2)(x2-2x+4)  =  0 

and  is  satisfied  if  x^  —  2x4-4=0 

as  well  as  if  x  +  2=0. 

The  solution  of  x^  —  2  x  +  4=0 

X  =  1  ±  V^. 


gives 


.-.  the  roots  are  i,  —  2,  1  ±  V-  3,  i  (—  1  ±  V—  3). 

(2)    Solve  x'-10x'  +  35x^-o0x-\-24:  =  0. 
Take  the  square  root  of  the  left  side. 

x'-10x^-\-S5x'—50x-{-24:\^  —  5x±5 


2x' 


10a;H35a:2 
10x^  +  25x^ 


2x'-10x-]-5 


10x^-50x  +  24: 
lOa^  — 50rr-f  25 


-   1 


It  is  now  seen  that  if  1  were  added,  the  square  would  be  complete 
and  the  equation  would  be 

a;*  — 10a;«  + 35x2  — 50^  +  25  =  1. 

Extract  the  square  root,  and  the  result  is, 

x2-5x+6=  ±1. 
That  is,  x2  —  5  X  =  —  4,  or  —  6. 

4x2- () +  25  =  9,  orl, 

2x  — 5=  ±3,  or±l, 
2x=8,  2,  6,  or  4. 
.-.  X  =  4,  1,  3,  or  2. 


214  ALGEBRA. 

Exercise  89. 
Find  the  roots  of : 

1.  x'-{-7x^  =  S.  8.  (0^2- 9)2  =  3  + 11  (^^2  — 2). 

2.  x^  —  5x^-{-4.  =  0.  9.  cc«  +  14^3  +  24=:0. 

3.  37x^  —  9  =  4:x\  10.  19a;^4-216cc^  =  x. 

4.  16x^  =  17x'-l.  11.  x«  + 22 0^^  +  21=0. 

5.  32 ic^''  — 33x^  +  1  =  0.  12.  a;2»'  +  3:^'^  — 4  =  0. 

6.  (x^-2y=i(x'-i-12).  13.  4:z;^-20^^+23^2_^5a;=6. 


5a^2«      25      ^  1,3 


7.    cc^  — ^-=^  =  0.  14.    f7„  +  :^,-20  =  0. 


15.  0^^  — 4a;3  — 10x2  +  28ic  — 15  =  0. 

16.  x*  —  2x'  —  lSx~  +  Ux  =  —  24:. 

17.  108x''  =  20a:(9:r2-l)  — 51a:2  +  7. 

18.  (x2-l)(x2-2)  +  (a^2_3^^^^2_4^)^^4_|.5^ 

Problems  Involving  Quadratics. 

236.  Problems  which  involve  quadratic  equations  have 
apparently  two  solutions,  as  a  quadratic  has  two  roots. 
Sometimes  both  will  be  solutions  :  but  generally  one  only 
will  be  a  solution,  and  the  other  be  inconsistent  with  the 
conditions  of  the  problem.  No  difficulty  will  be  found 
in  selecting  the  result  which  belongs  to  the  problem,  and 
sometimes  a  change  may  be  made  in  the  statement  of  a 
problem  so  as  to  form  a  new  problem  corresponding  to  the 
solution  which  was  inapplicable  to  the  original  problem. 


QUADRATIC    EQUATIONS.  216 

(1)  The  sum  of  the  squares  of  two  consecutive  numbers  is 

481.     Find  the  numbers. 

Let  X  =  one  number, 

and  x+  1  =  the  other. 

Then  ic2  +  (x  +  1)2  =  481, 

or  2x2  +  2x4-1  =  481. 

The  solution  of  which  gives,  x  =  15,  or  —  16. 

The  positive  root  15  gives  for  the  numbers,  15  and  16. 

The  negative  root  —  16  is  inapplicable  to  the  problem,  as  consecu- 
tive numbers  are  understood  to  be  integers  which  follow  one  another 
in  the  common  scale,  1,  2,  3,  4 

(2)  What  is  the  price  of  eggs  per  dozen  when  2  more  in  a 

shilling's  worth  lowers  the  price  1  penny  per  dozen  ? 

Let  X  =  the  number  of  eggs  for  a  shilling. 

Then,  -  =  the  cost  of  1  egg  in  shillings, 

12 

and  —  =  the  cost  of  1  dozen  in  shillings. 

But,  if  X  +  2  =  the  number  of  eggs  for  a  shilling, 

12 
■  ^  =  the  cost  of  1  dozen  in  shillings. 

X  "T"  J 

12         12  1    ,,  ^  .        ,      ^       ^Mi-     V 

•••  —  -  ^rp^  =  12  (^  P^^^y  ^^^^S  TJ  of  a  shillmg). 

The  solution  of  which  gives  x  =  16,  or  —  18. 
And,  if  16  eggs  cost  a  shilling,  1  dozen  will  cost  j|  of  a  shilling, 
or  9  pence. 

Therefore,  the  price  of  the  eggs  is  9  pence  per  dozen. 

If  the  problem  is  changed  so  as  to  read :  What  is  the 
price  of  eggs  per  dozen  when  two  less  in  a  shilling's  worth 
raises  the  price  1  penny  per  dozen  ?  the  algebraic  state- 
ment will  be 

12        12_  J_. 
X  — 2       X  ~  12' 

The  solution  of  which  gives  x  =  18,  or  —  16. 

Hence,  the  number  18,  which  had  a  negative  sign  and  was  inappli- 
cable in  the  original  problem,  is  here  the  true  result. 


216  ALGEBRA. 


Exercise  90. 


1.  The  sum  of  the  squares  of  three  consecutive  numbers 

is  365.     Find  the  numbers. 

2.  Three  times  the  product  of  two  consecutive  numbers 

exceeds  four  times  their  sum  by  8.  Find  the 
numbers. 

3.  The  product  of  three  consecutive  numbers  is  equal  to 

ihree  times  the  middle  number.     Find  the  numbers. 

4.  A  boy  bought  a  number  of  apples  for  16  cents.     Had 

he  bought  4  more  for  the  same  money  he  would  have 
paid  -J-  of  a  cent  less  for  each  apple.  How  many  did 
he  buy  ? 

5.  For  building  108  rods  of  stone-wall,  6  days  less  would 

have  been  required  if  3  rods  more  a  day  had  been 
built.     How  many  rods  a  day  were  built  ? 

6.  A  merchant   bought   some   pieces  of   silk   for   $900. 

Had  he  bought  3  pieces  more  for  the  same  money 
he  would  have  paid  $15  less  for  each  piece.  How 
many  did  he  buy  ? 

7.  A  merchant  bought  some  pieces  of  cloth  for  $168.75. 

He  sold  the  cloth  for  $12  a  piece  and  gained  as 
much  as  1  piece  cost  him.  How  much  did  he  pay 
for  each  piece  ? 

8.  Find  the  price  of  eggs  per  score  when  10  more  in  62|- 

cents'  worth  lowers  the  price  31^  cents  per  hundred. 

9.  The  area  of  a  square  may  be  doubled  by  increasing 

its  length  by  6  inches  and  its  breadth  by  4  inches. 
Determine  its  side. 
10.    The  length  of  a  rectangular  field  exceeds  the  breadth 
by  1  yard,  and  the  area  is  3  acres.     Determine  its 
dimensions. 


QUADRATIC    EQUATIONS.  217 

11.  There  are  three  lines  of  which  two  are  each  f  of  the 

third,  and  the  sum  of  the  squares  described  on  them 
is  equal  to  a  square  yard.  Determine  the  lengths 
of  the  lines  in  inches. 

12.  A  grass  plot  9  yards  long  and  6  yards  broad  has  a 

path  round  it.  The  area  of  the  path  is  equal  to 
that  of  the  plot.     Determine  the  width  of  the  path. 

13.  Find  the  radius  of  a  circle  the  area  of  which  would  be 

doubled  by  increasing  its  radius  by  1  inch. 

14.  Divide  a  line  20  inches  long  into  two  parts  so  that  the 

rectangle  contained  by  the  whole  and  one  part  may 
be  equal  to  the  square  on  the  other  part. 

15.  A  can  do  some  work  in  9  hours  less  time  than  B  can 

do  it,  and  together  they  can  do  it  in  20  hours. 
How  long  will  it  take  each  alone  to  do  it  ? 

16.  A  vessel  which  has  two  pipes  can  be  filled  in  two 

hours  less  time  by  one  than  by  the  other,  and  by 
both  together  in  2  hours  55  minutes.  How  long 
will  it  take  each  pipe  alone  to  fill  the  vessel  ? 

17.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both 
together  in  1  hour  52  minutes  30  seconds.  How 
long  will  it  take  each  pipe  alone  to  fill  the  vessel  ? 

18.  An  iron  bar  weighs  36  pounds.     If  it  had  been  1  foot 

longer  each  foot  would  have  weighed  -J-  a  pound 
less.     Find  the  length  and  the  weight  per  foot. 

19.  A  number  is  expressed  by  two  digits,  the  second  of 

which  is  the  square  of  the  other,  and  when  54  is 
added  its  digits  are  interchanged.    Find  the  number. 

20.  Divide  35  into  two  parts  so  that  the  sum  of  the  two 

fractions  formed  by  dividing  each  part  by  the  other 
may  be  2^. 


218  ALGEBRA. 

21.  A  boat's  crew  row  3J  miles  down  a  river  and  back 

again  in  1  hour  40  minutes.  If  the  current  of  the 
river  is  2  miles  per  hour,  determine  their  rate  of 
rowing  in  still  water. 

22.  A  detachment  from  an  army  was  marching  in  regular 

column  with  5  men  more  in  depth  than  in  front. 
On  approaching  the  enemy  the  front  was  increased 
by  845  men,  and  the  whole  was  thus  drawn  up  in 
5  lines.     Find  the  number  of  men. 

23.  A  jockey  sold  a  horse  for  $144,  and  gained  as  much 

per  cent  as  the  horse  cost.    What  did  the  horse  cost  ? 

24.  A  merchant  expended  a  certain  sum  of  money  in  goods, 

which  he  sold  again  for  $24,  and  lost  as  much  per 
cent  as  the  goods  cost  him.  How  much  did  he  pay 
for  the  goods  ? 

25.  A  broker  bought  a  number  of  bank  shares  ($100  each), 

when  they  were  at  a  certain  per  cent  discount,  for 
$7500 ;  and  afterwards  when  they  were  at  the  same 
pej^  cent  premium,  sold  all  but  60  for  $5000.  How 
many  shares  did  he  buy,  and  at  what  price  ? 

26.  The  thickness  of  a  rectangular  solid  is  f  of  its  width, 

and  its  length  is  equal  to  the  sum  of  its  width  and 
thickness ;  also,  the  number  of  cubic  yards  in  its 
volume  added  to  the  number  of  linear  yards  in 
its  edges  is  |  of  the  number  of  square  yards  in 
its  surface.      Determine  its  dimensions. 

27.  If  a  carriage-wheel  16^  feet  round  took  1  second  more 

to  revolve,  the  rate  of  the  carriage  per  hour  would 
be  l-J  miles  less.  At  what  rate  is  the  carriage 
travelling  ? 


CHAPTER   XV. 
Simultaneous  Quadratic  Equations. 

237.  Quadratic  equations  involving  tivo  unknown  quan- 
tities require  different  methods  for  their  solution,  according 
to  the  form  of  the  equations. 

238.  Case  I.  When  from  one  of  the  equations  the  value 
of  one  of  the  unknown  quantities  can  be  found  in  terms  of 
the  other,  and  this  value  substituted  in  the  other  equation. 

Ex.    Solve:  Sx^  —  2xij  =  5\  (1) 

x-y  =  2         J  (2) 

Transpose  x  in  (2),  y  =  x  —  2. 

Substitute  in  (1),       3  x2  —  2  x  (x  —  2)  =  5. 

The  solution  of  which  gives  x  =  1  or  —  5. 

.-.  y  =  —  1  or  —  7. 

Special  methods  often  give  more  elegant  solutions  of  examples  than 
the  general  method  by  substitution. 

I.  When  equations  have  the  fornix  x  ±  ?/  =  a,  and  xy  =  b;  x^  ±  y^ 
=  a,  and  xy=^b;  or,  x±y  =  a,  and  x'^  •}-  y^  =  b. 

(1)    Solve:  a-|-?/  =  40|  (1) 

a;y  =  300    J  (2) 

Square  (1),  x'^  +  2xy  +  y^  =  1600.  (3) 

Multiply  (2)  by  4,  4xy  =  1200.  (4) 

Subtract  (4)  from  (3),    x^  -  2  xy -{■  y'^  =  400. 
Extract  root  of  each  side,  x  —  y  =  ±20.  (6) 

Add  (1)  and  (5),  2  x  =  60  or  20. 

.-.  X  =  30  or  10. 
Subtract  (5)  from  (1),  2  y  =  20  or  60. 

.-.  y  =  10  or  30. 


220  ALGEBRA. 

(2)    Solve:  x  —  y=^4:\  (1) 

x'  +  7/  =  ^0J  (2) 

Square  (1),  x^-2xy  +  ?/  =  le.  (3) 

Subtract  (2)  from  (3),  —2xy=—  24.  (4) 

Subtract  (4)  from  (2),         x^  +  2  x?/  +  y'^  =  64. 
Extract  the  root,  x-{-  y=  ±S.  (5) 

By  combining  (5)  and  (1),  x  =  6  or  —  2. 


y  =  2  or  —  6. 


(^)  ^^^^^^       l+rlo  1 


(1) 

(2) 

(3) 

(4) 

(5) 


II.   When  one  equation  may  be  simplified  by  dividing  it  by  the  other. 


1 

x'' 

+? 

41 

~400j 

Square  (1), 

1  +  2       1  ^  81 . 
x^      xy      y^      400 

Subtract  (2)  from  (3) 

2        40 
xy      400 

Subtract  (4)  from  (2), 

1        2        1  _    1  ^ 
x2      xy      2/2      400 

Extract  the  root, 

X      y      =^20 

By  combining  (1)  and 

(5), 

X  =  4  or  5. 
y  =  5  or  4. 

Solve :                      X' 

^  +  2/^  =  911 
+  2/  =  7      J 

(1) 

X 

(2) 

Divide  (1)  by  (2), 

x2-xy  +  2/2=13. 

(3) 

Square  (2), 

x2+2xy  +  2/2  =  49. 

(4) 

Subtract  (3)  from  (4), 

3x?/  =  36. 

Divide  by  —  3, 

-xy=-  12. 

(5) 

Add  (5)  and  (3). 

x2  — 2x?/+  ?/2=  1. 

Extract  the  root, 

x-y=±\. 

(6) 

By  combining  (6)  and 

(2)S\                     X  =  4  or  3. 
2/  =  3  or  4. 

SIMULTANt:OUS    QUADRATIC    EQUATIONS. 


221 


Exercise  91. 


Solve  : 

1.  x-\-y  =  13\ 
xy  =  36      J 

2.  x  +  y  =  29^ 
xy  =  WO    J 

3.  x  —  y  =  19\ 
xy  =  66      J 

4.  X  —  ?/  =  45"l 
xy  =  250    J 

5.  x  —  y  =  10      \ 

x'-\-?f  =  17Sj 

6.  x  —  y  =  U      \ 
x^-\-f  =  A36j 

7.  x  +  y  =  12      I 


11.  x-\-y  =  49        1 
x'-\-y'  =  16Slj 

12.  x3+?/^=3411 
a:  +  y  =  ll      J 


13.  0:^  +  7/^  =  1008 1 

14.  x^—if=9S 


'-y'=9S\ 

-y  =  2    J 


15.    a:3-?/^  =  279\ 


16.  x  —  3y  =  l  \ 

17.  4y  =  5a«4-l 


y  =  5x-^l    1 
,r„  =  33-x'J 


8. 

'-  +  '-  = 

X      y 

3 
'4 

-  +  -- 

5 
~16, 

9. 

X      y 

=  5 

i  +  -  = 

=  13 

10.    7a;2  — 8:r2/  =  159| 
5a:  +  2y  =  7         J 


18.    i-i  =  3 

x^      f 


X9.    i-i  =  2^ 

a?       y 

x^      2/ 


20.    x^  —  2xy  —  2/  =  l 

x-^y  =  2 


} 


222  ALGEBRA. 

239.   Case  II.   When  each  of  the  two  equations  is  homo- 
geneous  and  of  the  second  degree. 

Ex.    Solve:  2if  —  4.xij-\-^x^  =  ll\  (1) 

if  —  x^  =  U  J  (2) 

Let        y  =  vx,  and  substitute  vx  for  y  in  both  equations. 
From  (1),  2  v'^x'^  —  4  ux2  +  3  x^  =  17. 

17 


■'■  ^       2  u2  -  4 1,  +  3 
From  (2),  v^x^  —  x'^=  16. 

17  16 


Equate  the  values  of  x^, 


2»2  —  4t)  +  3      u2_i 
32 1)2  -  64  u  +  48  =  17  u2  -  17, 
15i52-64tJ=-65. 


The  solution  gives, 

5       13 
.  =  -or-. 

Substitute  the  value  of  v  in 

-^' 

then, 

x2  =  9or^- 

.•.x=±3or±|» 

o 

and 

■         „         ,13 

y  =  vx  =  ±  ()  or  ±  — 

o 

Exercise  92. 
Solve : 

1.    £c2  +  :r?/  +  2/  =  74    1  4.  a;2-42/'  — 9=0  \ 

2^2  +  2x?/  +  y2  =  73j  ^^j^27/-3  =  0S 


3a;2+82/2==14      J  xy  +  ^f- 

x^  —  xy-}-y^  =  21'\  6.    cc^  +  ccy  +  2 t/^  =  44 \ 

2r^-2a;2/  =  -15  J  2«2-^y  +  2/2^1(3  J 


SIMULTANEOUS    QUADRATIC    EQUATIONS.  223 

7.  x^  +  xy  —  lo  =  0\  9.    2x^-\-3xy-\-y^  =  70\ 
x?/  —  f  —  2  =  0   J  6x^-}-xi/  —  i/  =  o0    J 

8.  x^  —  x7j-\-i/  =  7  1         10.    x^  —  xy  —  if  =  h         \ 

240.  Case  III.  When  the  two  equations  are  symmetrical 
with  respect  to  x  and  y;  that  is,  when  they  have  x  and  y 
similarly  involved  in  them. 

Thus,  the  expressions  2 x^  +  3 x"^y2  -^2y^,  2x?/  —  3x  —  3y  +  l, 
X*  —  3  x^y  —  3  xy2  +  yt  are  symmetrical  expressions. 

(1)    Solve:  x^-^f  =  r^xy\  (1) 

a:  +  7/  =  12        J  (2) 

Put  M  +  u  f or  X,  and  m  —  u  f or  y,  in  (1)  and  (2). 

(1)  becomes        (w  +  w)^  +  (m  —  r)^  =  18  (u  +  r)  (a  —  »), 

or  1*8  +  3  wu2  =  9  (u2  —  1)2).  (8) 

(2)  becomes  (w  +  t?)  +  (m  —  u)  =  12, 
or  2  M  =  12. 

.-.  w  =  6. 
Substitute  6  for  u  in  (3). 


(3)  becomes 
whence, 

and 

216  +  18 1)2  =  9  (36  -  »2), 

t)2=4. 

.•.tJ=±2, 

.-.  X  =  M  +  w  =  6  ±  2  =  8  or  4, 
y=M  —  »  =  6=F2=:4or8. 

(2)    Solve: 

aj  +  y  =  8        1                                 (1) 
a;<-[-2/^  =  706j                                 (2) 

Put  M  +  u  for  X,  and  w  —  u  for  y  in  (1)  and  (2). 

(1)  becomes  (w  +  v)  +  (w  —  u)  =  8. 

.-.  u  =  4. 

(2)  becomes  m*  +  6  m'-'u2  +  u*  =  353.  (8) 
Substitute  4  for  m  in  (3), 

256  +  96  r2  +  v*  =  353, 

or,  1)4  +  961)2  =  97.  (4) 

The  solution  of  (4)  gives  d  =  ±  1  or  ±  V— 97. 

Taking  the  possible  values  of  r,  x  =  5  or  3,  and  y  =  3  or  5. 


224  ALGEBRA. 

Exercise  93. 

Solve : 

1.  4.x7/  =  96  —  xy\  4.    4.(x-{-ij)  =  Sxi/         1 
x-\-y  =  6            J  x  +  2j-{-x^-{-f  =  26j 

2.  x''-\-f  =  lS  —  x  —  y\  5.    4tx^  +  xi/-i-4i/^  =  5S'\ 
xy  =  6                          J  50^24.5^2^35  J 

3.  2(x'  +  f)  =  5x7n  6.    xy(x-^y)  =  301 
4:(x  —  y)  =  xy      J  x^-\-i/^  =  35       J 

241.  The  preceding  cases  are  general  methods  for  the  solution  of 
equations  which  belong  to  the  kinds  referred  to ;  often,  however,  in 
the  solution  of  these  and  other  kinds  of  simultaneous  equations  in- 
volving quadratics,  a  little  ingenuity  will  suggest  some  step  by  which 
the  roots  may  easily  be  found. 

Exercise  94. 

Solve  : 

1.  x~-y  =  l  \  8.    x—y  =  l      \ 
^'  +  ^y  +  2/'=13j  x^-\^f^  =  ^\ 

2.  x'^-{-xy  =  3o\  9.    x^-^4.xy  =  3    \ 
xy-f=.Q    J  ^xy^y'  =  2i\ 

3.  xy  —  12  =  0\  10.    x'^  —  xy-{-y~  =  ^S\ 
x—2y=n   J  x—y—^=0        ) 

4.  xy  —  l=^0    \  11.    x''-{-3xy-{-y^  =  l    \ 
x^  +  f=^bOi  3:c2+:ry+3/=13  j 

5.  2x  —  5y  =  9        \  12.    x''—2xy-\-3y''=l^\ 
x'-xy  +  y'  =  7j  x^  +  xy-y'=^        ) 


6.  x  —  y  =  9    \  13.    x-\-y  =  a  } 
a;y  +  8  =  0)                                          4:xy  —  a'  =  —  U^) 

7.  5x  —  7y  =  0  "j  14.    x  —  y  =  l    ^ 


SIMULTANEOUS    QUADRATIC    EQUATIONS. 


225 


15.   x2  +  9a-y  =  340 
Ixy 


ry  =  340| 


16.  x-\-y  =  Q)      \ 

17.  Sxy-\-2x-}-i/  =  ^S5\ 
Sx  —  2y  =  0  j 

18.  x  —  i/  =  l      \ 

19.  0^3  +  2/^  =  2728         ■) 


a 


} 


20.  X  +  y: 

^  +  / 

21.  3^2  — 2/^  =  0 
3x^  — 4xy4-5y' 

ic-f-?/      cc  —  ?/      10 


-.1 

x  —  y      x-\-y       3    V 
a^  +  ,/  =  45  J 


23.    -  +  i  =  5 

X      2/ 


17 


x  +  1   '   y  +  l       12J 

24.  cc^  —  .r //  +  //-  =  7         ) 

a-^+.ry'+y=i33) 

25.  x-\-y  =  4.      I 


26.    a:"" —?/''= --3 


y  =  a     ) 


27.  a:^  — a;?/  =  «2_|_^2| 
xi/  —  i/^  =  2ab       ) 

28.  x'—7f  =  Aab\ 
xy^a'  —  h"^    ) 

29.  xy  =  0  \ 
x^-{-if  =  lQ>\ 

30.  a;2  +  j:2/  +  r  =  37      > 
x*  +  a:y  +  y  =  481) 

31.  £c^^aic  +  ^2/l 
f  =  ay-\-hx) 


32.    a;  —  ?/  — 2  =  0  | 

15(x2-y2)  =  16a:y) 


x-\-y  ,  x  —  y      89 

33.    — ^^  +  — -^^  =  — 

a^  —  y      x-\-  y      40 

6a;  =  20y  +  9 


} 


X   ,    ?/ 
34.    -  +  f  = 


^  +  ^  =  4 
a;       y 


35.  ic- +  ?/ =  7 -h  ir//    I 
a:3-|-7/8=6a'//— 1  ) 

36.  ir^  — //  =  3093') 
a:  —  y^3  ) 


+  7/^  =  82) 

37.  f(.x-l)-|(a^  +  l)(2/-l)  =  -ll) 
i(y  +  2)=i(x  +  2)  > 

38.  100^2+ 15.T//  =  3(?Z>  —  2an 
10/+ir).ry  =  3aZ>  — 2^^  ) 


226  ALGEBRA. 


Exercise  95. 

1.  If  the  length  and  breadth  of  a  rectangle  were  each  in- 

creased by  1,  the  area  would  be  48 ;  if  they  were 
each  diminished  by  1,  the  area  would  be  24.  Find 
the  length  and  breadth. 

2.  The  sum  of  the  squares  of  the  two  digits  of  a  number 

is  25,  and  the  product  of  the  digits  is  12.  Find  the 
number. 

3.  The  sum,  the  product,  and  the  difference  of  the  squares 

of  two  numbers  are  all  equal.     Find  the  numbers. 

Note.     Represent  the  numbers  hy  x  +  y  and  x  —  y,  respectively. 

4.  The  difference  of  two  numbers  is  f  of  the  greater,  and 

the  sum  of  their  squares  is  356.  What  are  the 
numbers  ? 

5.  The  numerator  and  denominator  of   one   fraction  are 

each  greater  by  1  than  those  of  another,  and  the 
sum  of  the  two  fractions  is  ly^^ ;  if  the  numerators 
were  interchanged  the  sum  of  the  fractions  would 
be  1-J.     Find  the  fractions. 

6.  A  man  starts  from  the  foot  of  a  mountain  to  walk  to  its 

summit.  His  rate  of  walking  during  the  second  half 
of  the  distance  is  ^  mile  per  hour  less  than  his  rate 
during  the  first  half,  and  he  reaches  the  summit  in 
5^  hours.  He  descends  in  3f  hours,  by  walking  1 
mile  more  per  hour  than  during  the  first  half  of  the 
ascent.  Find  the  distance  to  the  top  and  the  rates 
of  walking. 

Note.    Let  2x=  the  distance,  and  y  miles  per  hour  =  the  rate  at  first. 

XX  2  X 

Then  -  H =  5^  hours,  and  — —-  =  3f  hours, 

y     y-\  y+i 


SIMULTANEOUS    QUADRATIC    EQUATIONS.  227 

7.  The  sum  of  two  numbers  which,  are  formed  by  the 

same  two  digits  in  reverse  order  is  f|  of  their 
difference;  and  the  difference  of  the  squares  of  the 
numbers  is  3960.     Determine  the  numbers. 

8.  The  hypotenuse  of  a  right  triangle  is  20,  and  the  area 

of  the  triangle  is  96.     Determine  the  legs. 

Note.     The  square  on  the  hypotenuse  =  sum  of  the  squares  on 
the  legs ;  and  the  area  of  a  right  triangle  =  i  product  of  legs. 

9.  Two  boys  run  in  opposite  directions  round  a  rectan- 

gular field  the  area  of  which  is  an  acre ;  they  start 
from  one  corner  and  meet  13  yards  from  the  oppo- 
site corner;  and  the  rate  of  one  is  |  of  the  rate  of 
the  other.     Determine  the  dimensions  of  the  field. 

10.  A,  in  running  a  race  with  B  to  a  post  and  back,  met 

him  10  yards  from  the  post.  To  make  it  a  dead 
heat,  B  must  have  increased  his  rate  from  this  point 
41^  yards  per  minute;  and  if,  without  changing  his 
pace,  he  had  turned  back  on  meeting  A,  he  would 
have  come  in  4  seconds  after  him.  How  far  was  it 
to  the  post  ? 

11.  The  fore  wheel  of  a  carriage  turns  in  a  mile  132  times 

more  than  the  hind  wheel ;  but  if  the  circumferences 
were  each  increased  by  2  feet,  it  would  turn  only 
88  times  more.     Find  the  circumference  of  each. 

12.  A  person  has  $6500,  which  he  divides  into  two  parts 

and  loans  at  different  rates  of  interest,  so  that  the 
two  parts  produce  equal  returns.  If  the  first  part 
had  been  loaned  at  the  second  rate  of  interest,  it 
would  have  produced  $180 ;  and  if  the  second  part 
had  been  loaned  at  the  first  rate  of  interest,  it  would 
hav«  produced  $245.     Find  the  rates  of  interest. 


CHAPTER  XYI. 

Simple  Indeterminate  Equations. 

242.  If  a  single  equation  be  given  which,  contains  two 
unknown  numbers,  and  no  other  condition  be  imposed, 
the  number  of  its  solutions  is  U7ilimited;  for,  if  any  value 
be  assigned  to  one  of  the  unknown  numbers,  a  corresjpond- 
ing  value  may  be  found  for  the  other.  Such  an  equation 
is  said  to  be  indeterminate. 

243.  The  values  of  the  unknown  numbers  in  an  inde- 
terminate equation  are  dependent  upon  each  other;  so  that, 
though  they  are  unlimited  in  number,  they  are  confined  to 
a  particular  range. 

This  range  may  be  still  further  limited  by  requiring 
these  values  to  satisfy  some  given  condition;  as,  for 
instance,  that  they  shall  be  positive  integers. 

244.  The  method  of  solving  an  indeterminate  equation 
in  positive  integers  is  as  follows : 

(1)    Solve  3  a? -[-4?/ =  22,  in  positive  integers. 

Transpose,  3  x  =  22  —  4  ?/, 

....  =  7-^  +  1^, 

the  quotient  being  written  as  a  mixed  expression. 


Since  the  values  of  x  and  y  are  to  be  integral,  x  +  y  —  7  will  be 

begral,   and  hen( 
form  of  a  fraction. 


X  —  v 
integral,   and  hence,  — z-^  will  be  integral,  though  written  in  the 
o 


1—7/ 

Let  — ^  =  m,  an  integer ; 


SIMPLE    INDETERMINATE    EQUATIONS.  229 

Then  l  —  y=Sm, 

.:  y  =  1  —  3m. 
Substitute  this  value  of  y  in  the  original  equation, 
3x  +  4-12?/i  =  22, 

.-.  X  =  6  +  4  m. 

The  equation  y  =  l  —  Srn  shows  that  m  in  respect  to  y  may  be  0, 
or  have  any  negative  value,  but  cannot  have  a  positive  value. 

The  equation  x  =  6  +  4  m  shows  that  m  in  respect  to  x  may  be  0, 
but  cannot  have  a  negative  value  greater  than  1. 


.-.  m  may  be  0  or  —  1, 

and  then 

x  =  6,  y  =  l; 

or 

X  =  2,  y  =  4. 

(2)    Solve  5x  — 14?/ =11,  in  positive  integers. 

Transpose,  5  x  =  11  +  14  y, 

.•.x  =  2  +  2^  +  iJt.U?, 

1  +  4v 
.•.x-2y-2  =  -^^' 
o 

1  +  4y 
.'.  — ; —  must  be  integral, 
o 

Now,  if  — - —  be  put  =  m,  then  y  =  — j —  ,  a  fraction  in  form. 

To  avoid  this  difficulty,  it  is  necessary  in  some  way  to  make 

the  coefficient  of  y  equal  to  unity.     Since  — r—   is  integral,  any 

1  +  42/ 
multiple  of  — — ^  is  integral.     Multiply,  then,  by  such  a  number  as 

will  make  the  coefficient  of  y  greater  by  1  than  some  multiple  of  the 
denominator.     In  this  case,  multiply  by  4.     Then 

4  +  16y       0.4  +  2/..,        , 

— -  or  3  ?/  H —  is  mtegral. 

0  o 

4  +  y 
.-.  — r-^  =  m,  an  mteger ; 

.-.  y—  bm  —  4. 
Since    x=  \{\\  +  \^y),  from  the  original  equation, 

.-.  x=  14  m  — 9. 
Here  it  is  obvious  that  m  may  have  any  positive  value^  and 

x=5,  19,  33 

2/=l,  6,  11 


230  ALGEBRA. 

The  required  multiplier  can  always  be  found  when  the  coefficients 
are  prime  to  each  other,  and  it  is  best  to  divide  the  original  equation 
by  the  smaller  of  the  two  coefficients,  in  order  to  have  the  multiplier 
as  small  as  possible. 

246.  The  necessity  for  a  multiplier  may  often  be  obviated 
by  a  little  ingenuity.     Thus, 

The  equation        4  y  =  29  —  7  x    may  be  put  in  the  form  of 
4y  =  29  —  Sx+x, 

...2/=7-2x  +  ^' 

•    in  which  the  fraction  is  of  the  required  form. 
The  equation        5x=  18+13?/ 

gives  x=3  +  iiv  +  ^-^^-^-^^' 

1  +  V 
in  which  - — —^  is  of  the  required  form. 

246.  It  will  be  seen  from  (1)  and  (2),  §  244,  that  when  only 
positive  integers  are  required,  the  number  of  solutions  will 
be  limited  or  unlimited  according  as  the  sign  connecting 
X  and  ?/  is  positive  or  negative. 

(1)    Find  the  least  number  that  when  divided  by  14  and  5 
will  give  remainders  1  and  3  respectively. 

If  N  represents  the  number,  then 

=  X,  and  — z —  =  v. 

14  '  5 

.-.  JV^=14x  +  l,  andJV^=5y  +  3. 

.-.  14x+l  =  52/  +  3. 

52/  =  14x  — 2, 

5y=15x  — 2  — X. 

2  +  x 


Let 


Km=l, 


.'.  y  = 

3x-      ^    . 

2  +  x 
5 

:  wi,  an  integer ; 

.-.  z  = 

5m -2, 

y  = 

:  ^(14x  — 2),  from  < 

original  equation. 

.-.  y  = 

14  m  -  6. 

X  = 

:  3,  and  y=S. 

.:N  = 

:14x+l  =  62/+3 

=  43. 

Arts. 

SIMPLE    INDETERMINATE    EQUATIONS.  231 

(2)    Solve  5x-{-6y  =  S0,  so  that  x  may  be  a  multiple  of  y, 
and  both  positive. 


Let 

x=  my. 

Then 

{om  +  e)y  =  30. 

30 

id 

30  m 

If  m  =  2, 

x  =  Shy=n- 

If  m  =  3, 

x  =  U,  y=U. 

(3)    Solve  14 x  + 22// =  71,  in  positive  integers. 

If  we  multiply  the  fraction  by  7  and  reduce, 
the  result  is  —  4  ?/  +  |, 

a  form  which  shows  there  can  be  no  imtegral  solution. 

There  can  be  no  integral  solution  of  ax  ±  6y  =  c,  if  a  and  h  have  a 
common  factor  not  common  also  to  c  ;  for,  if  d  be  a  factor  of  a  and 
also  of  6,  but  not  of  c,  the  equation  may  be  written, 

vidx  ±  ndy  =  c, 

or  mx±ny  =  -^3k  fraction. 

a 


Exercise  96. 

Solve 

in  positive 

integers  : 

1. 

2x-\-lly: 

=  49. 

5. 

2. 

7x  +  3y  = 

^40. 

6. 

3. 

5x  +  7y  = 

=  53. 

7. 

4. 

x-\-10y  = 

:29. 

8. 

5.    Sx-\-Sy  =  61. 
Sx-\-5y  =  97. 
16a: +  72/  =  110. 
7x-\-10y  =  206. 

Solve  in  least  positive  integers : 

9.    12x  —  7y  =  l.  12.    2Sx  —  9y  =  929. 

10.  5a;  — 17?/  =  23.  13.    23 a;  —  33 ?/ =  43. 

11.  23?/  — 13a^  =  3.  14.   555x  —  22y  =  7S. 


232  ALGEBRA. 

15.  How  many  fractions  are  there  with  denominators  12 

and  18  whose  sum  is  f|  ? 

16.  What  is  the  least  number  which,  when  divided  by  3 

and  5,  leaves  remainders  2  and  3  respectively  ? 

17.  A  person  counting  a  basket  of  eggs,  which  he  knows 

are  between  50  and  60,  finds  that  when  he  counts 
them  3  at  a  time  there  are  2  over;  but  when  he 
counts  them  5  at  a  time  there  are  4  over.  How 
many  are  there  in  all  ? 

18.  A  person  bought  40  animals,  consisting  of  pigs,  geese, 

and  chickens,  for  $40.  The  pigs  cost  $5  apiece,  the 
geese  $1,  and  the  chickens  25  cents  each.  Find  the 
number  he  bought  of  each. 

19.  Find  the  least  multiple  of  7  which,  when  divided  by  2, 

3,  4,  5,  6,  leaves  in  each  case  1  for  a  remainder. 

20.  In  how  many  ways  may  100  be  divided  into  two  parts, 

one  of  which  shall  be  a  multiple  of  7  and  the  other 
of  9? 

21.  Solve  18x  —  5y  =  70  so  that  y  may  be  a  multiple  of  x, 

and  both  positive. 

22.  Solve  Sec +  12?/ =  23  so  that  x  and  y  may  be  positive, 

and  their  sum  an  integer. 

23.  Divide  70  into  three  parts  which  shall  give  integral 

quotients  when  divided  by  6,  7,  8,  respectively,  and 
the  sum  of  the  quotients  shall  be  10. 

24.  Divide  200  into  three  parts  which  shall  give  integral 

quotients  when  divided  by  5,  7,  11,  respectively,  and 
the  sum  of  the  quotients  shall  be  20. 

25.  A  number  consisting  of  three   digits,   of  which  the 

middle  one  is  4,  has  the  digits  in  the  units'  and 
hundreds'  places  interchanged  by  adding  792.  Find 
the  number. 


SIMPLE    INDETERMINATE    EQUATIONS.  233 

26.  Some  men  earning  each  $2.50  a  day,  and  some  women 

earning  each  $1.75  a  day,  receive  all  together  for 
their  daily  wages  $44.75.  Determine  the  number 
of  men  and  the  number  of  women. 

27.  A  wishes. to  pay  B  a  debt  of  £1  12.9.,  but  has  only  half- 

crowns  in  his  pocket,  while  B  has  only  4-penny  pieces. 
How  may  they  settle  the  matter  most  simply  ? 

28.  Show  that  323  a;  — 527  v/  =  1000  cannot  be  satisfied  by 

integral  values  of  x  and  y. 

29.  A  farmer  buys  oxen,  sheep,  and  hens.      The  whole 

number  bought  is  100,  and  the  whole  price  £100. 
If  the  oxen  cost  £5,  the  sheep  £1,  and  the  hens 
\s.  each,  how  many  of  each  did  he  buy? 

30.  A  number  of  lengths  3  feet,  5  feet,  and  8  feet  are  cut; 

how  may  48  of  them  be  taken  so  as  to  measure  175 
feet  all  together  ? 

31.  A  field  containing  an  integral  number  of  acres  less 

than  10  is  divided  into  8  lots  of  one  size,  and  7  of  4 
times  that  size,  and  has  also  a  road  passing  through 
it  containing  1300  square  yards.  Find  the  size  of 
the  lots  in  square  yards. 

32.  Two  wheels  are  to  be  made,  the  circumference  of  one 

of  which  is  to  be  a  multiple  of  the  other.  What 
circumferences  may  be  taken  so  that  when  the  first 
has  gone  round  three  times  and  the  other  five,  the 
difference  in  the  length  of  rope  coiled  on  them  may 
be  17  feet  ? 

33.  In  how  many  ways  can  a  person  pay  a  sum  of  £15 

in  half-crowns,  shillings,  and  sixpences,  so  that  the 
number  of  shillings  and  sixpences  together  shall  be 
equal  to  the  number  of  half-crowns  ? 


CHAPTEE   XVII. 

Inequalities. 

247.  Expressions  containing  any  given  letter  will  have 
their  values  changed  when  different  values  are  assigned  to 
that  letter ;  and  of  two  such  expressions,  one  may  be  for 
some  values  of  the  letter  larger  than  the  other,  for  other 
values  of  the  letter  smaller  than  the  other. 

Thus,  1  +  X  +  x^  will  be  greater  than  1  —  x  +  x'^  for  all  positive 
values  of  x,  but  less  for  all  negative  values  of  x. 

.  248.  One  expression,  however,  may  be  so  related  to 
another  that,  whatever  values  may  be  given  to  the  letter, 
it  cannot  be  greater  than  the  other. 

Thus,  2  X  cannot  be  greater  than  x^  +  1,  whatever  value  be  given 
to  X. 

249.  For  finding  whether  this  relation  holds  between  two 
expressions,  the  following  is  a  fundamental  proposition  : 

If  a  and  h  are  unequal,  a^-\-b^^2 ah. 

For,  (a  — 6)2  must  be  positive,  whatever  the  values  of  a  and  h. 
That  is  (a  — 6)2>0, 

or  a2_2a&  + 6-2>0; 

.-.  a2  + 52-^2  a6. 

250.  The  principles  applied  to  the  solution  of  equations 
may  be  applied  to  inequalities,  except  that  if  each  side  of 
an  equality  have  its  sign  changed,  the  inequality  will  be 
reversed. 

Thus,  if  a  >  6,  then  —  a  will  be  <  —  &. 


INEQUALITIES.  235 

(1)  If  a  and  h  be  positive,  show  that  a^  +  ^^  is  >  a%  +  ab^. 

if  (dividing  each  side  by  a  +  &), 

a2  — a6  +  62>a6, 
if  a2  +  &2>2a6. 

But  a2  +  62is>2a6.  §249. 

.-.  a8  +  &3>a26  +  a52. 

(2)  Show  that  a'-\-h'^-{-  c^  is  >  a^*  +  ac  +  be. 

Now,  a2  +  62is>2a6, 

a2  +  c2  is  >  2  ac,  §249. 

62  +  c2is>2  6c. 
By  adding,  2 a^  +  2 62  +  2 c2  is> 2 a6  +  2  oc  +  2 6c. 

.-.  a2  +  62  +  c2  is>  a6  +  ac  +  6c. 


Exercise  97. 
Show  that,  the  letters  being  unequal  and  positive : 

1.    a^-\-Un^>2b(a-\-b).  2.    a^b -\- ah^  i^>  2  a^bK 

3.  {a'  +  Z»2)  (a^  -f  ^,4^^  is  >  (««  +  ^y^)^. 

4.  a^^,  _p  ^2^  _|_  «^2  _|.  yi^  _j_  ^^2  _|.  5^  ig  ;>  g  ^^^ 

5.  The  sum  of  any  fraction  and  its  reciprocal  is  >  2. 

6.  If  x^=a^-\-b~,  i/^e^-\-(P,  xy  is  not  <  ac-\-bd,  or  ac?+&c. 

7.  ah-\-iw^bc<{ci-^b  —  cf-\-{a-^G  —  b)--{-(b^c  —  af. 

8.  Which  is  the  greater,  {a?-{-W)  {(?-^d^)  or  {ac^bdYl 

9.  Which  is  the  greater,  m^  -f  m  or  m^  + 1  ? 

10.    Which  is  the  greater,  a^—b^  or  4  a^  {(i—b)  when  a  is>J  ? 


11 


.    Which  is  the  greater,  '^-j  +  \—  or  y/a-\-\lb? 


12.   Which  is  the  greater,     T    or  — r—:  ? 

2  a  +  6 


13.    Which  is  the  greater,  B~f~72^^i:"f"7^ 


a  ,    b        1,1 

r2  +  ~2  0^  T  +  ~ 
b^      of       b       a 


CHAPTER  XVIII. 

Theory  of  Exponents. 

251.  The  expression  a%  when  n  is  a  positive  integer,  has 
been  defined  as  the  product  of  71  equal  factors  each  equal 
to  a.  §  24. 

And  it  has  been  shown  that  a"'  X  «"  =  «"*  +  ".  §  66. 

That  a""  -^-  o^«  =  a'"-",  if  m  is  greater  than  n-,  §  93. 

or »  if  m  is  less  than  n.  §  94. 

And  that  («-)«  =  tt"-".  •  §  199. 

Also,  it  is  true  that  a"  X  ^"  =  (fx^)" ;  for 
(^aby  =  ah  taken  n  times  as  a  factor, 

=  a  taken  n  times  as  a  factor  X  h  taken  n  times  as  a 
factor  =  6t"  X  y^. 

252.  Likewise,  -\/«,  when  w  is  a  positive  integer,  has  been 
defined  as  one  of  the  n  equal  factors  of  a  (§  203)  ;  so  that  if 
\la  be  taken  n  times  as  a  factor,  the  resulting  product  is  a; 
that  is,  {^aY  =  a. 

Again,  the  expression -7^  means  that  a  is  to  be  raised 
to  the  77ith  power,  and  the  nth  root  of  the  result  obtained. 

And  the  expression  (-7^)"*  means  that  the  nth  root  of  a 
is  to  be  taken,  and  the  result  raised  to  the  mth  power. 

It  will  thus  be  seen  that  any  proposition  relating  to  roots 
and  powers  may  be  expressed  by  this  method  of  notation. 
It  is,  however,  found  convenient  to  adopt  another  method  of 
notation,  in  which  fractional  and  negative  exponents  are 
used. 


THEORY    OF    EXPONENTS.  237 

253.  The  meaning  of  a  fractional  exponent  is  at  once  sug- 
gested, by  observing  that  the  division  of  an  exponent^  when 
the  resulting  quotient  is  integral,  is  equivalent  to  extracting 
a  root.  Thus,  a^  is  the  square  root  of  a^,  and  3,  the  expo- 
nent of  a^,  is  obtained  by  dividing  the  exponent  of  a^  by  2. 

If  this  division  be  indicated  only,  the  square  root  of  a^ 
will  be  denoted  by  ai,  in  which  the  denominator  denotes  the 
root,  and  the  numerator  the  poiver.  If  the  same  meaning 
be  given  to  an  exponent  when  the  division  does  not  give  an 
integral  quotient,  a^  will  represent  the  square  root  of  the 
cube  of  a ;  and,  in  general,  a",  the  nth  root  of  the  mth  power 
of  a.  This,  then,  is  the  meaning  that  will  be  assigned  to  a 
fractional  exponent,  so  that  in  a  fractional  exponent 

254.  The  numerator  will  indicate  a  power,  and  the  denom- 
inator a  root.  ^ 

255.  The  meaning  of  a  negative  exponent  is  suggested 
by  observing  that  in  a  series  of  descending  powers  of  a, 

a" a*,  a*,  a^,  a^,  a^, 

the  subtraction  of  1  from  the  exponent  is  equivalent  to 
dividing  by  a;    and  if  the   operation  be   continued,  the 

result  is 

a^,  a~^,  a-^,  a"^,  a"* a"". 

Then  a*'  =  -  =  1 ;    a~^  =  1  -j-  «  =  - ; 

a  a 

a   '  a^'  a" 

This,  then,  is  the  meaning  that  will  be  assigned  to  a 
negative  exponent,  so  that, 

256.  A  number  with  a  7iegative  exponent  will  denote  the 
reciprocal  of  the  number  with  the  corresponding  positive 
exponent. 


238  ALGEBRA. 

It  may  be  easily  shown  that  the  laws  which  apply  to 
positive  integral  exponents  apply  also  to  fractional  and 
negative  exponents. 

mm  m 

257.    To  show  that  a^Xb^=  {ah) « : 

^  —         n  n,     • 


=  •V(a&)'», 

m 

=  (ab) »     (by  definition). 
Ill  1 

Likewise     a»  X  6»  X  c«  =  (a5c)»,  and  so  on. 

1  1  1 


268.    To  show  that  («"*)«  =  «""*: 


•  ±  t 

Let  X  =  (a"*)". 

1 
Then  a:"  =  a"*,  and  x"»«  =  a. 

j_ 
.*.  X  =  a""». 
1  1 
But  X  =  (a™)",    (by  supposition). 

1  i        JL 

.-.  (a'")"  =  a*"". 

269.    To  show  that  a""  X  «""  =  a"*"" : 
Now  a"*  X  «-»  =  a»*  X  — » 

or  = if  m  <  n,  §  94. 

=  a-("-™)     (by  definition), 
=  a"*— «. 

260.    In  like  manner  the  same  laws  may  be  shown  to 
apply  in  every  case. 


THEORY    OF   EXPONENTS.  239 

261.   Hence,  whether  m  and  n  are  integral  or  fractional, 
positive  or  negative: 

I.    a"»  X  a"  =  «'"+".  III.    (a"*)"  =  a"*". 

II.    a"*^- a"  =  «"«-".  IV.    a'^Xh^'^iaby. 

Exercise  98. 
Express  with  fractional  exponents  : 

1.  V^;   \^?;    (VJ)^;   \/^;   -J/^^;    (v'^/;   ^^l 

2.  \/i^«;   ^iy?;    ^^W;    5  s/^^b^. 

Express  with  radical  signs : 

3.  aij  aib^j  Axhj-i-^  Sxhj-i. 

Express  with  positive  exponents  : 

Write  in  the  form  of  integral  expressions : 

Sxy     _z_      a         (?        x-^     x-^ 
^'    "?■'    ^*'    be'    a%-^'    y-V    y\' 

Simplify  : 

6.  ahXa^;  bhXbl;  c^Xchy  diX  d^. 

7.  miXm-i;  n^Xn-i"^-,  a^Xa^)  a^  X  a'i. 

8.  ai  X  Va ;  c"*  X  Vc ;  y^X^y;  xiX  V^. 

9.  a^>ic  X  a-iftci;  a35*c-i  X  aJ^-ic^c?. 

10.    iciz/MXa;-8y-i;5J-i;  aii/i^i  X  a;-*?/-*^-*. 


11. 


aiX«-iXa-JX«-»;   (f)*X  (j)'x  (^,J 


240  ALGEBRA. 

3,— 

12.  ai-T-  ai;  ct  -^  d',  m^^  -i-  ni;  a^ -h  Va  . 

13.  (ft«)^-^(a«)§;   (c-^)§;   (m-hy-,   (ni)'';   (xi)*. 

14.  (i?-2)-^;   W~^;(a^"W~^;   (aiXa^)-il 

15.  (4a-§)-i:  (276-^)-?;  (64ci«)-i;   (32c-^«)i 

-•  (w)~ '  (il^)" '  (^'O-^  (fit)-. 

262.  The  laws  that  apply  to  the  exponents  of  simple 
expressions  also  apply  to  the  exponents  of  compound  ex- 
pressions. 

(1)    Multiply  2/1 +2/^ +  2/^  +  1  by  ^i-1. 

yi  +  yh  +  yk+1 
yk-l 


y  -\- y^ -\-  yh -h  yi 
—  yl  —  yh  —  y\  —  \ 


—  1.     Ans. 


(2)    Divide  xl-^x^  — 12  by  ic*  —  3. 


4x^-12 

4  X*  —  12  X*  +  4.   Ans. 


,,  ,..  ,  Exercise  99. 

Multiply : 

1.  x^P-\-  xPyP  +  ifp  by  cc^p  —  x^y^  +  i/^. 

2.  a;""*-"  —  ?/"  by  x"  +  ?/"'"-". 

3.  cc!  — 2a:;i  +  l  by  x^  —  1. 

4.  8a»  +  4al^»*  +  5aW  +  9Mby  2«f  — 51 


THEORY    OF    EXPONENTS. 


241 


5.1  +  ab-^  +  a%-^  by  1  -  ah-^  +  a'b-^ 

6.  ^2^-2  +  2  +  a-%^  by  a^^-^  —  2  —  a-^^*^. 

7.  4rr-3  +  3x-2  +  2a;-i  +  l  by  x'^  — aj-^  +  l. 

Divide : 

8.  x^  —  if""  by  ic"  —  y". 

9.  x-\-ij-\-z  —  ?>xhjH\\)Yx^-\-y^-\-zK 

10.  x  +  ?/byxt  —  a;iy*  +  a:2y*  —  ^*y^  +  ^« 

11.  X V'  +  2  +  x-hf  by  x?/-i  +  x-^//. 

12.  a-^  +  a-2^-2  +  Zi-''  by  a-^  -  ^-^^-^  +  h-\ 

Find  the  squares  of : 

13.  4a^>-^;  ai  — ^i;  a-^a-^\  2a%^  —  a-\h\, 

If  a  =  4,  ^  =  2,  c  =  1,  find  the  values  of  : 

14.  a\h;  oab-';  2  (ab)^',  a-ib-'d;  12a-%-\ 

15.  Expand  (a5-^»i)^   (2x-'  +  a:/;   {ab-'-by-y. 

Extract  the  square  root  of : 

16.  9x-^  — 18a:-V+15a:-V-6ic-y  +  y2. 

Extract  the  cube  root  of : 

17.  8a:^  +  12.r2-30.x-3o  +  4ox-^  +  27.x-2-27a:-^ 

Resolve  into  prime  factors  with  fractional  exponents : 

18.  -^12?  \/72j  V'OGj  V64  ;    and  find  their  product. 

Simplify : 

1  1 

19.  ;(a;*'*)^X(x")-2j3a-2.  20.    (x^«"  X  a;-^^)3«-2. 

21.3  («i  +  hhf  -  4  (ai  +  ^i)  («i  —  Z'i)  +  («i  —  2by. 

22.     {(a'")"*"^}^^.  24.    [J  (a-"*)-" 5'^]*^-^ [J  (a'")" 5-^]-^. 


23.    I—  )^l  —  1      .  25.     ^,,,_,,^^,,,_,, 


242  ALGEBRA. 


Radical  Expressions. 

263.  An  indicated  root  that  cannot  be  exactly  obtained  is 
called  a  surd.  An  indicated  root  that  can  be  exactly  ob- 
tained is  said  to  have  the  form  of  a  surd. 

264.  The  required  root  shows  the  order  of  a  surd ;  and 
surds  are  named  quadratic,  cubic,  biquadratic,  according  as 
the  second,  third,  or  fourth  roots  are  required. 

265.  The  product  of  a  rational  factor  and  a  surd  factor 
is  called  a  mixed  surd;  as,  3  V2,  b  \la. 

266.  When  there  is  no  rational  factor  outside  of  the 
radical  sign,  the  surd  is  said  to  be  entire;  as,   V2,  Va. 

267.  Since  yla  X  V^  X  v  c  =  \labc,  the  product  of  two  or 
more  surds  of  the  same  order  will  be  a  radical  expression 
of  the  same  order  consisting  of  the  product  of  the  numbers 
under  the  radical  signs. 

268.  In  like  manner,  \/a%  =  Va^  X'\/b  =  a  ^Ib.     That  is, 
A  factor  under  the  radical  sign  whose  root  can  be  taken, 

may,  by  having  the  root  taken,  be  removed  from  under  the 
radical  sign. 

269.  Conversely,  since  a^/b=  \la^b, 

A  factor  outside  the  radical  sign  may  be  raised  to  the 
corresponding  power  and  placed  under  it. 


Again  :  \|=  \^  ^h^^J^""'' 


and  yj^=y]^  =  yjabxl,==l^b. 


RADICAL    EXPRESSIONS.  243 

270.  A  surd  is  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  integral  and  as  small  as  ijossible. 

271.  Surds  which,  when  reduced  to  the  simplest  form, 
have  the  same  surd  factor,  are  said  to  be  similar. 

Simplify : 

V50j      -^108;      ^7^^     >JJ;      i^/J^,;      ^296352. 

(1)  V50  =  V25  X  2  =  5  V2. 

(2)  ^Jm>  =  ^^27  X  4  =  3  \^. 

(3)  -v/TxV  :=  -^7x22/2  X  y6  =  yKllxy^. 


*  '   \(26V      \16Mc«       26c 


(6)   V206342. 


23 

296352 

22 

37044 

32 

9261 

3 

1029 

7 

343 

7 

49 

Hence,  296352  =  2«  x  3^  x  7^, 
.-.  -5^296352  =  \^3  X  \^38  X  \^, 
=  7X3X2  V22, 
=  42V4.   Ans. 


In  simplifying  numerical  expressions  under  the  radical  sign,  the 
method  employed  in  (6)  may  be  used  with  advantage  when  the  factor 
whose  root  can  be  taken  is  not  readily  determined  by  inspection. 

Exercise  100. 
Express  as  entire  surds  : 

1.  3V5;    3V2I;    5V32;    a%\lhc',   x^^. 

2.  32/^^',    2x^xy',    a^^a^^',    Sc^Vabc;    5abcy/abP\ 


;.  fV^;  leVfH;  (^+2/)\':74:^ 


a^-\-2xy  +  f 


244  ALGEBRA. 

Express  as  mixed  surds  : 


4.     \lx\fz',    -sl^a^;    V546iV/;    V24 ;    ^126  a^d\ 


5.     VlOOOa;     V160a;y;    Vl087/iV%    V1372a^^Z^^«. 


Simplify : 


7.  2V80«56V;  7V396a;;  9V81a;y^;  5V726. 

8.  V|;    ViU;    V3i;  fV90|;  2^^. 


3 /2a-?/      3,—     a  W  ^,  /Sa^Z'a; 


cif 

12       2       /xy\  /  ^'  y    /'^'\/£!^\^ 
^^-   V5'  Vrroi'  V  ^V  V^yy  '  V'  A  ^  / 

11.  (ao;)  X  (^2;^)^ ;   (2  6^2^,4^  X  (^'V)* ;  5  (3  a^^V)  X  (a'b-*f)^. 

12.  Show  that  V20,  V45,  V|^  are  similar  surds. 

13.  Show  that  2ya%'^,  ySb^,  i^^  ^^^  similar  surds. 

14.  If  V2  =  1.414213,  find  the  values  of 

272.    Surds    of   the  same   order   may   be    compared   by 

expressing  them  as  entire  surds. 

Ex.    Compare  fVT  and  f  VlO. 

|Vio  =  V^. 

V^  =  V^,  and  VJ^  =  V^. 
As  V^  is  greater  than  V-^^^^^^,  f VlO  is  greater  than  fV?. 


RADICAL    EXPRESSIONS.  245 

273.  The  product  or  quotient  of  two  surds  of  the  saine 
order  may  be  obtained  by  taking  the  product  or  quotient 
of  the  rational  factors  and  the  surd  factors  separately. 

(1)  2V5  X  5  V7=  10  V35. 

(2)  9V5-v-3V7  =  3V|  =  3V||  =  fV35. 


Exercise   101. 

1.  Which  is  the  greater  SVl  or  2Vl5? 

2.  Arrange  in  order  of  magnitude  9  VS,  GVT,  SVlO. 

3,-  S,—  3,— 

3.  Arrange  in  order  of  magnitude  4V4,  3V5,  5v3. 

4.  Multiply  3  V2  by  4V6  ;  f  VlO  by  /^  VlS. 

5.  Multiply  5  V^  by  f  Vi62  ;  ^-v^i  by  2\/2. 

6.  Divide  2  V5  by  3  Vi5 ;  f  V2l  by  ^^  V^. 

7.  Simplify  §  V3  X  |  V5  -f-  f  V2. 

^     -,.      ...    2ViO  ^^  7V48        4Vi5 

8.  Simplify  — p=  X  — p=  -7- prr, 

3V27      5Vl4       15V21 

9.  Simplify  2  ^/4X  5  \/32-f-\^108. 

274.  The  order  of  a  surd  may  be  changed  by  changing 
the  power  of  the  expression  under  the  radical  sign.     Thus, 

Vo  =  V25  ;  y/c  =  Vc^. 
Conversely,  v26  =  V5 ;  v c^  =  V c  j 

mn, —  mi— 

or,  in  general,  Vc"  =  Vc 

In  this  way,  surds  of  different  orders  may  be  reduced  to 
the  same  order^  and  may  then  be  compared,  multiplied,  or 
divided. 


246  ALGEBRA. 

(1)  To  compare  V2  and  v3. 

V2  =  2^  =  2^  =  V^  =  ^/S; 
^^=  3i  =  3^  =  ^  =  -\j9. 
.-.  Vi  is  greater  than  V2. 

(2)  To  multiply  V4a  by  V6cc. 

-^  =  (4 a)^  =  (4  a)§  =  V'(4 a)2  =  'V^16a2  ; 
Ve^  =  (6  x)^  =  (6  x)^  =  \/(6x)3  =  \/216x8. 
.-.  ^te  X  ^/6x  =  \^16a2  x  \/216x3. 


=  Vl6a2x  216x3, 
=  'v'2%2  X  23  X  38x3. 
=  ^^26  X  2  X  33a2x3, 


(3)    To  divide  ^Sa  by  -J6b. 

^/S^z=  (3 a)^  =  (3 a)§  =  \/(3a)2  =  y/da^  ; 
V66  =  (6  b)^  =  (6  6)^  =  v'(66)"3  =  '(721063. 


...  ^3^ -V66  =  ^/972- ^2-166-3=^^: 


24&3     A^23x363 


23  X   35^263  1     6 


26  X  3666        66 


=  —  V1944  a263.   Ans. 


Exercise  102. 
Arrange  in  order  of  magnitude  : 
1.    2^,  3V2,  1^4.  3.    2^,  3^7,  4V2. 


RADICAL   EXPRESSIONS.  247 

Simplify : 

6.  3  (4  aby  -^  (2  a%)h ;  (2  a^b^  X  (a'b')i  -r-  (a%y, 

7.  (2 a^»)i  X  (3 aZ*2)i  -f.  (5 a^>3) J ;  4  Vl2  -i-  2  V3. 

'■(?)' K?)' -(.0- 

9.    (7V2-5V6-3V84-4V20)X3V2. 


10.  Vttf)'XV(||)^    ^/{4.abyx^(2a'by. 

11.  (  ^^)3  X  (  V^y ;  ai&-3clc^-i  -f-  aift-ftc-He^tt. 

275.  In  the  addition  or  subtraction  of  surds,  each  surd 
must  be  reduced  to  its  simplest  form ;  and,  if  the  resulting 
surds  are  similar. 

Add  the  rational  factors,  and  to  their  sum  annex  the 
common  surd  factor. 

If  the  resulting  surds  are  not  similar, 
Connect  them  with  their  proper  signs. 

276.  Operations  with  surds  will  be  more  easily  per- 
formed if  the  arithmetical  numbers  contained  in  the  surds 
are  expressed  in  their  prime  factors,  and  if  fractional  expo- 
nents are  used  instead  of  radical  signs. 

(1)    Simplify  V27  +  V48  +  Vl47. 

V27  =  (3>^)2  =  3  X  3*  =  3  V3  ; 
Vis  =  (2-*  X  3)^  =  22  X  3^  =  4  X  3^  =  4V3  ; 
Vl47  =  (72  X  3)i  =  7  X  3^  =  7  Vs. 
.-.  V27  +  Vis  +  Vli7  =  (3  +  4  +  7)  V3  =  14  V3.   Ans. 


248  ALGEBRA. 

(2)    Simplify  2  \/320- 3  \/40. 

2\/320  =  2(26x  5)3  =  2  X  22  X  5^=  8\/5; 
3  \/40  =  3  (23  X  5)3  =  3  X  2  X  53  =  6  \/5 
.•.2\^320-3\^=8V^-6\^  =  2'V^5.  Ans. 


(3)  Find  the  square  root  of  V81. 

The  square  root  of  \/8T  =  (81^)^  =  81*  =  (3*)* 
=  3§  =  (32)3  =  -v^. 

(4)  Find  the  cube  of  |  ^. 

The  cube  of  i  v'2  =  (i)3  X  (2*)3  =  ^  x  2^  ==  i  V2. 

Exercise  103. 
Simplify : 

1.  V27-I-2V48  +  3V108;  3V1000  +  4V50  +  12V288. 

2.  \/l28  +  \^686+\/i6;  7^/54  +  3\^T6+ ^^432. 


3.  I2V72  — 3V128;  7V81  — 3V1029. 

4.  2V3  +  3Vli-V5i;  2V|+V60-Vl5-Vf. 


6.  V4 a^^  +  V25 ab^ -(a  — 5b)  ^ab. 

7.  c'\fa^^  —  a^abh^-\-b'\la%^c^. 

8.  2\/40  +  3'v^i08  +  \^500-\/320  — 2'V^1372. 

9.  (2-^3^)^  (3^3)1  10.    {^^ifJ)  (V27)i 

11.  (^8T)i;    (\^512)^;    (\^256)-i;   ^^16;   \/27. 

12.  '-\/4;   \/36;  '-732;   \/243;   \/i25;   -1^49. 

13.  \^«:    -^9^^   V'le^;    \^32^. 


RADICAL    EXPRESSIONS.  249 

14.  (-^8)^  (\^27)^   (\/64)3;  (^ly. 

15.  (a^a)-'-,  (x^x)-h-,  (;/V^)^;   (a-^^'^')-l 

Expand  by  the  method  explained  in  §  201  : 

16.  (^/a  +  ^by-,  (-v^m^+V^)^  {'\f^  —  2\fhy. 

17.  (2a^-iVa)«;  (2^x^-hfy-.   {y~^^J' 

"■(^-i7.)'(i-:-^)'(-0- 

Find  the  square  root  of : 

21.  x''^ +  Qar^"i/'-{- 11  x^y^-{- 6x^^1/^  +  1/*^. 

22.  l  +  4x-i-2x-2  — 4x-i  +  25a-J  — 24x-«  +  16a;-2. 

3 
277.   If  we  wish  to  find  the  approximate  value  of  ~1=* 

it  will  be  less  labor  to  multiply  first  both  numerator  and 
denominator  by  a  factor  that  will  render  the  denominator 
ratio7ial;  in  this  case  by  V2.     Thus, 

3  3V2  3V2 


V2       V2  X  V2         2 


278.  It  is  easy  to  rationalize  the  denominator  of  a  frac- 
tion when  that  denominator  is  a  binomial  involving  only 
quadratic  surds.  The  factor  required  will  consist  of  the 
same  terms  as  the  given  denominator,  but  with  a  different 


250  ALGEBRA. 

sign  between  them.     Thus,  will  have  its  denomi- 

nator  rationalized  by  multiplying  both  terms  of  the  fraction 
hJQ  —  2'^|l.     For, 

7-3V5^(7~3V5)(6-2V5) 
6  +  2  V5  ~"  (6  +  2  V5)  (6  —  2  V5) 

72-32V5      9_2^^ 


16  2 

279.  By  two  operations  the  denominator  of  a  fraction 
may  be  rationalized  when  that  denominator  consists  of 
three  quadratic  surds. 

Thus,  if  the  denominator  is  VC  +  V3  —  V2,  both  terms  of 
the  fraction  may  be  multiplied  by  VG  —  V3  +  V2.  The 
resulting  denominator  will  be  6  —  5  +  2  VG  =  1  +  2  V6  ; 
and  if  both  terms  of  the  resulting  fraction  be  multiplied 
by  1  —  2  V6,  the  denominator  will  become  1  —  24  =  —  23. 


Exercise  104. 

Find  equivalent  fractions  with  rational  denominators,  for 
the  following : 

3  7  4— V2  6 


1. 


2. 


V7  + V5'2V5-V6'  I+V2'  5  — 2V6 
a  a-\-h      2x  —  ^xy 


Vi  —  Vc     a  —  V^     ^xy  —  2y 


Find  the  approximate  values  of : 

_2_^  __1 7V5        7  +  2VlO_ 

V3'  V5-V2'  V7  +  V3'7-2VlO 


radical  expressions.  251 

Imaginary  Expressions. 

280.  All  imaginary  square  roots  may  be  reduced  to  one 
form.  

V— a;2=  Va:^  X  (— l)  =  icV— 1. 

V^i  =  VaX(— 1)  =  ah  V^. 

281.  V— 1  means  an  expression  which,  when  multiplied 
by  itself,  produces  —  1.     Therefore, 

(V^)^=-ij_         _  _  _ 

(V-l)-^=:(V-l)'xV-l  =  ~lV-l  =  -V-l; 
(V^)^  =  (V^)^X(V^)^  =  (-1)X(-1)  =  1; 

and  so  on.     So  that  the  successive  powers  of  V—  1  form 
the  repeating  series,  +  V—  1,  —  1,  —  V— 1,4-1. 

(1)   Multiply  1  +  V^  by  1-  —  V^. 

1-V^=1-2V-1. 
(1  +  2  V^)  (1-2  V^)  =  1  -  4  (-  1)  =  5. 


(2)    Divide  V—  ab  by  v—  b. 

and  V^   =6W^. 

=  Vo. 


V-6        64V^ 
Exercise  105. 


Multiply : 

1.  4  +  V^by4-V^;  Vs  — 2  V^  by  V3-f  2  V^. 

2,  V54  by  V—  2 ;  a  V—  bhj  x  V —  i/. 


252  ALGEBRA. 


3.   V— a  + V— ^  by  V— ^  — V— ^;  a V—  a^b*  by  V—  a*b\ 


4.  V-lObyV-2;  2V3-6V--5  by  4V3  -  V-5. 

Divide : 

5.  xsZ—l  by  ?/ V—  1 ;  1  by  V—  1 ;  a  by  a^yJ—1. 


6.  V— 12byV-3j   Vl5byV-5;  V-5byV-20. 


Square  Root  of  a  Binomial  Surd. 

282.  The  product  or  quotient  of  two  dissimilar  quadratic 
surds  will  be  a  quadratic  surd.     Thus, 

^ab  X  ^labc  =  a^  Vc ; 
^abc  -r-  y/ab  =  Vc. 

For  every  quadratic  surd,  when  simpUfied,  will  have  under  the 
radical  sign  one  or  more  factors  raised  only  to  the  first  power ;  and 
two  surds  which  are  dissimilar  cannot  have  all  these  factors  alike. 

Hence,  their  product  or  quotient  will  have  at  least  one  factor 
raised  only  to  the  first  power,  and  will  therefore  be  a  surd. 

283.  The  sum  or  difference  of  two  dissimilar  quadratic 
surds  cannot  be  a  rational  number^  nor  can  it  be  expressed 
as  a  single  surd. 

For  if  Va±  '^b  could  equal  a  rational  number  c,  we  should  have, 
by  squaring, 

a±2Va6+6=c2; 

that  is,  ±2  Va6  =  c^  —  a  —  b. 

Now,  as  the  right  side  of  this  equation  is  rational,  the  left  side 
would  be  rational ;  but,  by  §  282,  Vab  cannot  be  rational.  Therefore, 
Va  ±  Vft  cannot  be  rational. 

In  like  manner,  it  may  be  shown  that  Va  ±  Vb  cannot  be  ex- 
pressed as  a  single  surd  Vc. 


RADICAL    EXPRESSIONS.  253 

284,  A  quadratic  surd  cannot  equal  the  sum  of  a  rational 
number  and  a  surd. 

For,  if  Va  could  equal  c  +  Vft^  we  should  have,  by  squaring, 
a  =  c2  +  2cV6"+6, 
and,  by  transposing,  2c^/b=  a  —  b  —  c^. 

That  is,  a  surd  equal  to  a  rational  number,  which  is  impossible. 

286.  If  a-\-  y/b  =  x-\-  Vy,  then  a  will  equal  x  and  b  will 
equal  y. 

For,  by  transposing,  Vft  —  Vi7=  x  —  a ;  and  if  6  were  not  equal  to 
y,  the  difference  of  two  unequal  surds  would  be  rational,  which  by 
§  283  is  impossible. 

.-.  6  =  y,  and  a  =  x. 

In  like  manner,  if  a  —  V6  =  x  —  Vy,  a  will  equal  x  and  6  will 
equal  y. 

286.    To  extract  the  square  root  of  a  binomial  surd  a  -j-  V^. 


Let  Va+  V6  =  >^+Vy. 

Squaring,  a  +  V&  =  x  4-  2  Vxy  +  y. 

.\z  +  y  =  a,  and  2  Vxy  =  Vft!  §  285. 

From  these  two  equations  the  values  of  x  and  y  may  be  found. 

This  method  may  be  shortened  by  observing  that,  since 

\lb=^2\lxy,  

a  —  '/b  =  x  —  2^xy  +  y. 
By  taking  the  root,  \ja  —  Vi  =  Va?  —  Ny. 

.-.  ( V«  +  V^)  ( V^  -  V^i)  ==  (  V^  +  V^)  (  V^  -  V^). 
.'.yaF--b  =  x  —  y. 
And,  as  a  =  a:  +  y, 

the  values  of  x  and  y  may  be   found   by  addition  and 
subtraction. 


254  ALGEBRA. 

Ex.   Extract  the  square  root  of  7  -|-  4  V3. 


Let 

V^-j.V?/=  V7  +  4V3. 

Then 

%/^_V^=  V7-4V3 

By  multiplying, 

a;  -  y  =  V49  -  48. 

.\x^y=l. 

But 

x  +  y=7.     • 

.♦.  a;  =  4,  and  y  =  8. 

.'.\fx  +  \fy  =  2  +  ^/3. 

,' 

.V7  +  4V3=2  +  V3. 

Exercise  106. 
Extract  the  square  roots  of: 

1.  14  +  6V5.  6.   20  — 8V6.  11.  14  — 4V6. 

2.  17  +  4Vi5.  7.    9  — 6V2.  12.  38  — 12ViO. 

3.  10  +  2V2i.  8.    94-42V5.  13.  103-12Vri. 

4.  16  +  2^/55.  9.    13  — 2V3O.  14.  57-12715. 

5.  9  — 2Vi4.  10.    11  — 6V2.  15.  3^  — Vio. 


16.    2a  +  2\/a^  —  b^         18.    87  — 12V42. 


17.    a^  —  2b^/a^~h\  19.    (a-\'by  —  4:(a  —  b)^/ab. 

287.  A  root  may  often  be  obtained  by  inspection.  For 
this  purpose,  write  the  given  expression  in  the  form  a-\-2  \/b, 
and  determine  what  two  numbers  have  their  sum  equal  to 
a,  and  their  product  equal  to  b. 

(1)    Eind  by  inspection  the  square  root  of  18  +  2  V77. 

It  is  required  to  find  two  numbers  whose  sum  is  18  and  whose 
product  is  77  ;  and  these  are  evidently  11  and  7. 


Then  18  +  2  V77  f  H  +_7  +  2_Vll  +  7, 

=  (Vll  +  V7)2. 
That  is,  \^  +  Vt"  =  square  root  of  18  +  2  VtT. 


RADICAL   EXPBESSIONS.  255 

(2)   Find  by  inspection  the  square  root  of  75  — 12  V21. 

It  is  necessary  that  the  coeflBcient  of  the  surd  be  2  ;  therefore, 
75  —  12  V21  must  be  put  in  the  form  of 

75-2V62X  21  =  75-2V756. 

The  two  numbers  whose  sum  is  75  and  whose  product  is  756 
are  63  and  12. 

Then         75-2  V756  =  63  +  12-2  V63  +  12, 
=  (V63-Vl2)2. 

That  is,      V63  —  Vl2  =  the  square  root  of  75  —  12  ^^; 
or,  3  V7  —  2  Vs  =  the  square  root  of  75  — 12  V2T. 


Equations  containing  Kadicals. 

\,  An  equation  containing  a  single  radical  may  be 
solved  by  arranging  the  terms  so  as  to  have  the  radical 
alone  on  one  side,  and  then  raising  both  sides  to  a  power 
corresponding  to  the  order  of  the  radical. 


Ex.   Va^— 9  +  ic  =  9. 

Vic2  —  9  =  9  —  a;. 
By  squaring,  x^  —  9  =  81  —  18  x  +  «^ 

18a;  =  90. 
.-.  X  =  5. 

289.   If  two  radicals  are  involved,  two  steps  may  be 
necessary. 

Ex.  Va;  +  15  +  V^  =  15. 

Vx+15  4-Vx=15. 
By  squaring, 


X  +  15  +  2  VxM^TSx  +  X  =  225. 
By  transposing,  2  Vx^  +  15  x  =  210  —  2  «. 

By  dividing  by  2,         Vx^  +  15  x  =  105  —  x. 
By  squaring,  x^  + 15  x  =  11025  —  210  x  +  xK 

225  x=  11025. 
.-.  X  =  49. 


256 


ALGEBRA. 


Some  of  the  following  radical  equations  will  reduce  to 
simple  and  others  to  quadratic  equations. 

golyg .  Exercise  107. 


1.   Va^  — 5  =  2.  6.   VaJ  +  4  +  V2a^  — 1  =  6. 


2.    2V3x  +  4  — 0^  =  4.  7.   Vl3a;— 1  — V2a— 1=5. 


3.    3  — Vx-'  — l  =  2ic.  8.   V4  +  a;  +  Vir  =  3. 


4.    V3a^  — 2  =  2(rr-4).  9.   V25+a^  +  V25  — ic  =  8. 


5.   4a:  — 12Vic  =  16.  10.    x^  =  21  +  -Jx^  —  9. 


11.    2a;-V8a;3  +  264-2  =  0. 


12.   Va:  +  l  +  Vx  +  16  =  Va:  +  25. 


13.   ^2x-\-l~^Jx  +  4.  =  i^x-3. 


14.    \/x  +  3  +  ^x  +  S  =  5^x. 
15.   ^/3-i-x-{^'s/x■■ 


^x  —  2a-{-^x-\-2 


^^     8x±-Ux-£^^        ^^      V7rr^  +  4  +  2V3a;-l^^ 
3ic  — V4x  — a;2        ■  *     V7£c2  +  4  — 2V3ic-l 


21.   '\/(x  —  ay-]-2ab-{-b^  =  x  —  a-{-b. 


22.   V(a;  +  a)2  +  2a^»  +  ^.2=.^,_a_a;. 

^3.  viT3+vr"^=# 


RADICAL    EXPRESSIONS.  257 

24.  4ca^  —  S(xh-^  1)  (xh  —  2)  =  xh (10  —  3xi). 

25.  (a:3  — 2)(xi  — 4)  =  xS(a:§  — 1)2  — 12. 

26.  ic^  — 4xi  =  96.  28.    xh -{- 2  a^x'h  =  3 a. 

3  -       81 

27.  x  +  x-'  =  2.9.  29.    8lVx  +  f3  =  52a;. 

Va; 

290.   Equations  may  be  solved  with  respect  to  an  expres- 
sion in  the  same  manner  as  with  respect  to  a  letter. 

(1)    Solve  (x^ -xy-S(x^-x)  +  12  =  0, 

Consider  {x^  —  x)  as  the  unknown  number. 
Then  (x^  -  x)2  -  8  {x^  -  x)  =  -  12. 

Complete  the  square,  (x^  —  x)2  —  ( )  +  16  =  4. 
Extract  the  root,  (x^  —  x)  —  4  =  ±  2. 

x2  —  X  =  6  or  2. 
Complete  the  square,  4x2  —  ( )  +  1  =  25  or  9. 

Extract  the  root,  2x  —  l=±5or±3. 

2x  =  6,  -4,4,  -2. 
.-.  X  =  3,  -  2,  2,  -  1. 


(2)    Solve  5x-7x2  —  8V7a;2  —  5a;  +  l  =  8. 

Change  the  signs  and  annex  +  1  to  both  sides. 

7  x2  -  5x  +  1  +  8V7x2-5x+l  =  -  7. 


Solve  with  respect  to  V7x2  — 5x+  1. 
(7x2-5x+l)  +  8(7x2-5x+  1)^+16  =  9. 

(7x2- 5x4-1)^  +  4  =±3. 

(7x2 -5x+  i)i=:-i  or -7. 
Square,  7  x*  —  5  x  +  1  =  1  or  49. 

Transpose,  7  x^  —  5  x  =  0  or  48. 

From  7x2  —  5x  =  0,        x  =  Oorf; 
From  7  x2  —  5  X  =  48,      x  =  3  or  —  2f . 

Note.  In  verifying  the  values  of  x  in  the  original  equation,  it  is 
seen  that  the  value  of  V7  x2  —  5  x  +1  is  negative.  Thus,  by  putting 
0  for  X  the  equation  becomes  0  —  8  Vl  =  8 ;  and  by  taking  —  1  f or  Vl 
we  have  (-  8)  (-  1)  =  8 ;  that  is,  8  =  8. 


258  ALGEBRA. 


(3)    Solve  a;2^a;  +  l  +  -  +  ^  =  l. 


Arrange  as  follows :  (x^  +  —  j-\-(x  +  -j=0. 

By  adding  2  to  (  a^  +  "^)' 

1       /        1\® 
there  is  obtained         x^  +  2  +  —  =  I  x-{-  -)  • 

x^      \        x) 

Multiply  by  4  and  complete  the  square, 

4(aj  +  ^y+()  +  l  =  9. 

Extract  the  root,      2(^a:  +  -Wl  =  ±3. 

2^x  +  ^)  =  2or-4. 

»  +  -  =  1  or  -  2. 

X 

'  Multiply  by  X,  x'^  —  x  =  —  \,       and        x2  +  2  x  =  —  1. 

.-.  4x2-()+ 1= -3,  .•.x2  +  2a:+l  =  0. 

2  X  -  1  =  ±  V^,  X  +  1  =  0. 

.-.  x  =  i  (1  ±  V^).  .'.X--  1. 

291.  An  equation  like  that  of  (3)  which  will  remain 
unaltered  when  -  is  substituted  for  x,  is  called  a  reciprocal 
equation. 

It  will  be  found  that  every  reciprocal  equation  of  oM  degree  will 
be  divisible  by  x  —  1  or  cc  +  1  according  as  the  last  term  is  negative 
or  positive  ;  and  every  reciprocal  equation  of  even  degree  with  its  last 
term  negative  will  be  divisible  by  x^—  1.  In  every  case  the  equation 
resulting  from  the  division  will  be  reciprocal. 


RADICAL    EXPRESSIONS.  259 

(4)    Solve  x'-i-2x*  —  3x^  —  3x'-^2x-{-l=0. 

This  is  a  reciprocal  equation,  for,  if  ic-i  is  put  for  x,  the  equation 
becomes  x-^  +  2x-*  —  Sx-^  —  Sx-^  +  2x-^  +1  =  0,  which  multiplied 
by  x^  gives  1  +  2  x  —  3  x^  —  3  x^  +  2  x^  +  x^  =  0,  the  same  as  the 
original  equation. 

The  equation  may  be  written  (x^  +  1)  +  2 x(x3+ 1)  —  3x2(x  +  1)  =  0, 
which  is  obviously  divisible  by  x  +  1.     The  result  from  dividing  by 
X  +  1   is  x''  +  x3  -  4x2  4-  X  +  1  =  0,   or   (x*  +  1)  +  x  (x2  +  1)  =  4x2. 
By  adding  2  x^  to  (x*  +  1)  it  becomes  (x*  +  2  x^  +  1)  =  (x^  +  1)2. 
Then  (x2  +  1)2  +  x  (x2  +  1)  =  6  x2. 

Multiply  by  4  and  complete  the  square, 

4(x2  +  l)2+()  +  x2  =  25x2. 
Extract  the  root,  2  (x2  +  1)  +  x  =  ±  5  x 

Hence,  2  x2  +  2  =  4  x  or  —  6  x. 

By  simplifying,  x2  —  2  x  =  —  1 ;      and  x2  +  3  x  =  —  1, 
whence,  x  =  1  and  1;    whence,  x  =  i(— 3  ±  V5). 

Therefore,   including  the  root  —  1  obtained  from  the  factor 
X  +  1,  the  five  roots  are  —  1,  1,  1,  i  (—  3  ±  Vs). 

By  this  process  a  reciprocal  cubic  equation  may  be  reduced  to  a 
quadratic,  and  one  of  the  fifth  or  sixth  degree  to  a  biquadratic,  the 
solution  of  which  may  be  easily  efEected. 


Solve 


Exercise  108. 


1.    x^  —  3x  —  6y/x^  —  3x  — 3  +  2  =  0. 

3.  (2a^  —  3xy  —  2(2x'  —  3x)  =  15. 

4.  (ax  —  by-\-4:a(ax  —  ^)  =  "~7~* 

5.  S(2x'  —  x)  —  (2x^  —  x)h  =  2. 

6.  15x  —  3x'  +  4:(x^  —  5x  +  5)h  =  16. 

7.    a^+x-^  +  x  +  x-^  =  4:.  9.    x^  +  x-{-^(x^  +  x)h  =  l. 


8.    a;2+Va:2-7  =  19.  10.    (x  +  iy  +  (x  —  l)i  =  5. 


260  ALGEBRA. 


11.    x  —  l  =  2  +  2x-l  12.    V3a:  +  5  — V3a;  — 5  =  4. 

13.    {x'+l)  —  x(x'  +  l)  =  —  2x'. 


14.    2x^  —  2^2x^  —  5x  =  5(x  +  S). 


15.    x-}-2  —  4.x\/x  +  2  =  12x^. 


16.    's/2x-i-a+'\/2x  —  a  =  b. 


17.  -j9x'  +  21x-\-l-^/9x'  +  6x'}-l  =  3x, 

18.  x^  — 4a:?  +  iz;-i  +  4a;-3  =  — |. 

19.  (2a^  +  32/)2  — 2(2a:  +  32/)=:8-|^ 
x'-i/  =  21  j 


20.  0:;  +  ?/  + V^jhy  =  a"1  22.    x^ -{- y^ -{- x -{- y  =  4:8 
^  —  2/4"  ^^  —  y  =  ^  J  ic?/  =  12 

21.  ic*  —  «y +  7/^  =  13^  23.    ic2  +  ir^?/2^a2. 

24.  (x-?/)2-3((r-2/)  =  10| 
icy  —  3a7?/=54  J 

25.  Vic  —  V^/ =  ic^  (a;^  +  2/*)  1^ 
(x  +  2/)'  =  2(a:-2/)2       J 


0^2/  —  fic  + 1/)  =  54  J 


26. 

iCy  — (0^  +  2/): 

27.  a;  +  2/+V^  =  28  | 
ie^  +  2/'  +  ^2/  =  336/ 

28.  -~Sax  =  ^/4.a^  +  9ax^+^' 
a  4 

29.  (a!  +  l  +  aj-^)(cc  — l  +  x-i)  =  5J. 

30.  2(xh  —  l)-^  —  2(xi-'4:y'r=S(xi  —  2)-\ 


CHAPTER   XIX. 

Logarithms. 

In  the  common  system  of  notation  the  expression 
of  numbers  is  founded  on  their  relation  to  ten. 

Thus,  3854  indicates  that  this  number  contains  10^  three  times,  10^ 
eight  times,  10  five  times,  and  four  units. 

293.  In  this  system  a  number  is  represented  by  a  series 
of  different  powers  of  ten,  the  exponent  of  each  power 
being  integral.  But,  by  employing  fractional  exponents, 
any  number  may  be  represented  (approximately)  as  a  single 
power  of  10. 

294.  When  numbers  are  referred  in  this  way  to  10,  the 
exponents  of  the  powers  corresponding  to  them  are  called 
their  logarithms  to  the  base  10. 

For  brevity  the  word  "  logarithm  "  is  written  log. 

From  §  255  it  appears  that : 

10°=    1,  io-n=TV)    =0.1, 

10^=   10,  10-2(=T*Tr)    =0.01, 

10^  =  100,  10-«(=^^Vtt)  =  0.001, 

and  so  on.     Hence, 

log      1  =  0,  log  0.1     =-1, 

log    10  =  1,  log  0.01    =  —  2, 

log  100  =  2,  log  0.001  =  -  3, 

and  so  on. 


262  ALGEBRA. 

It  is  evident  that  the  logarithms  of  all  numbers  between 

1  and       10  will  be      0  +  a  fraction, 

10  and     100  will  be       1  -f  a  fraction, 

100  and  1000  will  be      2  +  a  fraction, 

1  and  0.1      will  be  —  1  +  a  fraction, 

0.1    and  0.01    will  be  —  2  +  a  fraction, 

0.01  and  0.001  will  be  -  3  +  a  fraction. 

295.  The  fractional  part  of  a  logarithm  cannot  be 
expressed  exactly  either  by  common  or  by  decimal  frac- 
tions :  but  decimals  may  be  obtained  for  these  fractional 
parts,  true  to  as  many  places  as  may  be  desired. 

If,  for  instance,  the  logarithm  of  2  be  required ;  log  2  may  be 
supposed  to  be  ^. 

Then  10*  =  2  ;  or,  by  raising  both  sides  to  the  third  power,  10  =  8, 
a  result  which  shows  that  ^  is  too  large. 

Suppose,  then,  log  2  =  y%.  Then  lO*''''  =  2,  or  by  raising  both  sides 
to  the  tenth  power,  lO^  =  2io.  That  is,  1000  =  1024,  a  result  which 
shows  that  -^^  is  too  small. 

Since  i  is  too  large  and  -f^  too  small,  log  2  lies  between  \  and  j% ; 
that  is,  between  0.33833  and  0.30000. 

In  supposing  log  2  to  be  ^,  the  error  of  the  result  is  i^^^  =  -f^  =  0.2. 
In  supposing  log  2  to  be  j\,  the  error  of  the  result  is  ^"-Y^V-^^  = 
f^2^%  =  —  0.024 ;  log  2,  therefore,  is  nearer  to  -j\  than  to  ^. 

The  difference  between  the  errors  is  0.2  —  (—  0.024)  =  0.224,  and  the 
difference  between  the  supposed  logarithms  is  0,33333  —  0.3  =  0.03333. 

The  last  error,  therefore,  in  the  supposed  logarithm  may  be  con- 
sidered to  be  approximately  2V?  of  0.03333  =  0,0035  nearly,  and  this 
added  to  0.3000  gives  0.3035,  a  result  a  little  too  large. 

By  shorter  methods  of  higher  mathematics,  the  logarithm  of  2  is 
known  to  be  0,3010300,  true  to  the  seventh  place. 

296.  The  logarithm  of  a  number  consists  of  two  parts, 
an  integral  part  and  a  fractional  part. 

Thus,  log  2  =  0.30103,  in  which  the  integral  part  is  0,  and  the  frac- 
tional part  is  .30103  ;  log  20  =  1.30103,  in  which  the  integral  part  is  1, 
and  the  fractional  part  is  .30103. 


LOGARITHMS.  263 

297.  The  integral  part  of  a  logarithm  is  called  the  char- 
acteristic ;  and  the  fractional  part  is  called  the  mantissa. 

298.  The  mantissa  is  always  made  positive.  Hence,  in 
the  case  of  numbers  less  than  1  whose  logarithms  are  nega- 
tive, the  logarithm  is  made  to  consist  of  a  negative  charac- 
teristic and  a  positive  mantissa. 

299.  When  a  logarithm  consists  of  a  negative  character- 
istic and  a  positive  mantissa,  it  is  usual  to  write  the  minus 
sign  over  the  characteristic,  or  else  to  add  10  to  the  char- 
acteristic and  to  indicate  the  subtraction  of  10  from  the 
resulting  logarithm. 

Thus,  log.  0.2  =  1.30103,  and  this  may  be  written  9.30103  —  10. 

300.  The  characteristic  of  a  logarithm  of  an  integral 
number y  or  of  a  mixed  number,  is  one  less  than  the  number 
of  integral  digits. 

Thus,  from  §  294,  log  1  =  0,  log  10  =  1,  log  100  =  2.  Hence,  the 
logarithms  of  all  numbers  from  1  to  10  (that  is,  of  all  numbers  con- 
sisting of  one  integral  digit),  will  have  0  for  a  characteristic  ;  and  the 
logarithms  of  all  numbers  from  10  to  100  (that  is,  of  all  numbers  con- 
sisting of  two  integral  digits),  will  have  1  for  a  characteristic ;  and  so 
on,  the  characteristic  increasing  by  1  for  each  increase  in  the  number 
of  digits,  and  therefore  always  being  1  less  than  that  number. 

301.  The  characteristic  of  a  logarithm  of  a  decimal  frac- 
tion is  negative,  and  is  equal  to  the  number  of  the  place  occu- 
pied by  the  first  significant  figure  of  the  decimal. 

Thus,  from  §  294,  log  0.1  =  -  1,  log.  0.01  =  -  2,  log.  0.001  =  -  3. 
Hence,  the  logarithms  of  all  numbers  from  0. 1  to  1  will  have  —  1  for 
a  characteristic  (the  mantissa  being  plus) ;  the  logarithms  of  all  num- 
bers from  0.01  to  0. 1  will  have  —  2  for  a  characteristic  ;  the  logarithms 
of  all  numbers  from  0.001  to  0.01  will  have  —  3  for  a  characteristic ; 
and  so  on,  the  characteristic  always  being  negative  and  equal  to  the 
number  of  the  place  occupied  by  the  first  significant  figure  of  the  decimal. 


264  ALGEBRA. 

302.  The  mantissa  of  a  logarithm  of  any  integral  nutn- 
ber  or  decimal  fraction  depends  only  upon  the  digits  of  the 
number,  and  is  unchanged  so  long  as  the  sequence  of  the 
digits  remains  the  same. 

For,  changing  the  position  of  the  decimal  point  in  a  number  is 
equivalent  to  multiplying  or  dividing  the  number  by  a  power  of  10. 
Its  logarithm,  therefore,  will  be  increased  or  diminished  by  the  expo- 
nent of  that  power  of  10 ;  and,  since  this  exponent  is  integral^  the 
mantissa  of  the  logarithm  will  be  unaffected. 
Thus,  if  2719G  =  104-4345, 

then  2719.6=  103-4345, 

27.196=  101-4345, 
2.7196=    100-4345, 
0.27196  =    109-4345-10, 

0.0027196  =  107-4345-10. 

303.  The  advantage  of  using  the  number  10  as  the  base 
of  a  system  of  logarithms  consists  in  the  fact  that  the 
mantissa  depends  only  on  the  sequence  of  digits,  and  the 
characteristic  on  the  position  of  the  decimal  point. 

304.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore. 

The  logarithm  of  a  product  is  the  sum  of  the  logarithms 
of  the  factors. 

Thus,  log  20  =  log  (2  X  10)  =  log  2  +  log  10, 

=  0.3010  +  1.0000  =  1.3010 ; 
log  2000  =  log  (2  X  1000)  =  log  2  +  log  1000. 

=  0.3010  +  3.0000  =  3.3010; 
log  0.2  =  log  (2  X  0.1)  =  log  2  +  log  0.1, 

=  0.3010  +  9.0000  -  10  =  9.3010  -  10, 
log  0.02  =  log  (2  X  0.01)  =  log  2  +  log  0.01, 

=  0.3010  +  8.0000  -  10  =  8.3010  -  10. 

Exercise  109. 

Given:  log  2  =  0.3010;  log  3  =  0.4771;  log  5  =  0.6990; 
log  7  =  0.8451. 

Find  the  logarithms  of  the  following  numbers  by  resolv- 


LOGARITHMS.  265 

ing  the  numbers  into  factors,  and  taking  the  sum  of  the 
logarithms  of  the  factors  : 

1.  log  6.  9.  log  25.  17.  log  0.021.  25.  log  2.1. 

2.  log  15.  10.  log  30.       18.  log  0.35.  26.  log  16. 

3.  log  21.  11.  log  42.  ,  19.  log  0.0035.  27.  log  0.056. 

4.  log  14.  12.  log  420.  20.  log  0.004.  28.  log  0.63. 
\<^log35.  13.  log  12.  21.  log  0.05.7  29.  log  1.75.^^ 

6.  log  9.        14.  log  60.       22.  log  12.5.         30.  log  105. 
3®.  log  8.        15.  log  75.       23.  log  1.25.;;       31.  log  0.0105. /t^' 
8.  log  49.      16.  log  7.5.      24.  log  37.5.         32.  log  1.05. 

306.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore, 

The  logarithm  of  a  power  of  a  number  is  equal  to  the 
logarithm  of  the  number  multiplied  by  the  exponent  of  the 
power. 

Thus,        log  57  =    7  X  log  5  =    7  X  0.6990  =  4.8930. 
log  311  =  11  X  log  3  =  11  X  0.4771  =  5.2481. 

306.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore, when  roots  are  expressed  by  fractional  indices, 

The  logarithm  of  a  root  of  a  niwiber  is  equal  to  the  loga^ 
rithm  of  the  number  multiplied  by  the  index  of  the  root. 

Thus,       log  2i        =  i  of  log  2  =  i  X  0.3010  =  0.0753. 
log  0.002*  =  i  of  (7.3010  -  10). 

The  expression  \  of  (7.3010  -  10) 

may  be  put  in  the  form  of  i  of  (27.3010  -  30)  which  =  9.1003  -  10 ; 
for,  since  20  —  20  =  0,  the  addition  of  20  to  the  7,  and  of  —  20  to  the 
—  10,  produces  no  change  in  the  vialue  of  the  logarithm. 

307.  In  simplifying  the  logarithm  of  a  root  the  equal  pos- 
itive and  negative  numbers  to  be  added  to  the  logarithm  must 
be  such  that  the  resulting  negative  number,  when  divided  by 
the  index  of  the  root,  shall  give  a  quotient  of  — 10. 


266  ALGEBRA. 

Exercise  110. 

Given:  log  2  =  0.3010;  log  3  =  0.4771;  log  5=0.6990 
log  7  =  0.8451. 

Find  logarithms  of  the  following : 


1. 

2\ 

6. 

5^ 

11. 

5i 

16. 

V. 

21. 

5i. 

2. 

51 

7. 

2k 

12. 

7A. 

17. 

5§. 

22. 

2V. 

3. 

r^^ 

8. 

5i 

13. 

2i  -v"^ 

18. 

3ft. 

23. 

51.^"^ 

4. 

3«. 

9. 

3i.^ 

v^ 

14. 

5i      . 
3?.  0/'' 

19. 

75. 

^ 

24. 

7V. 

5. 

v.  ^^ 

10. 

7i 

15. 

20. 

3i. 

25. 

211.  ^"^ 

308.  Since  logarithms  are  simply  exponents  (§  294), 
therefore, 

The  logarithTn  of  a  quotient  is  the  logarithm  of  the  divi- 
dend minus  the  logarithm  of  the  divisor. 

Thus,     log  I  ==  log  3 -log  2  =0.4771 -0.3010=      0.1761. 
log  I  =  log  2  —  log  3  =  0.3010  -  0.4771  =  -  0.1761. 

To  avoid  the  negative  logarithm  —0.1761,  we  subtract  the  entire 
logarithm  0.1761  from  10,  and  then  indicate  the  subtraction  of  10 
from  the  result. 

Thus,  —0.1761  =  9.8239—10. 

Hence,  log  §  =  9.8239  -  10. 

309.  The  remainder  obtained  by  subtracting  the  loga- 
rithm of  a  number  from  10  is  called  the  cologarithm  of 
the  number,  or  arithmetical  complement  of  the  logarithm 
of  the  number. 

Cologarithm  is  usually  denoted  by  colog,  and  is  most 
easily  found  by  beginning  ivith  the  characteristic  of  the  loga- 
rithm and  subtracting  each  figure  from  9  down  to  the  last 
significant  figure,  and  subtracting  that  figure  from  10. 

Thus,  log  7  =  0.8451;  and  colog  7  =  9.1549.  Colog  7  is  readily- 
found  by  subtracting,  mentally,  0  from  9,  8  from  9,  4  from  9,  5  from 
9,  1  from  10,  and  writing  the  resulting  figure  at  each  step. 


LOGARITHMS.  267 

310.  Since  colog  7  =  9.1549, 

and       log  I  =  log  1  -  log  7  =  0  -  0.8451  =  9.1549  - 10, 
it  is  evident  that. 

If  10  he  subtracted  from  the  cologarithm  of  a  number,  the 
result  is  the  logarithm  of  the  reciprocal  of  that  number. 

311.  Since  log  J  =  log  7  —  log  5, 

=  0.8451  -  0.6990  =  0.1461, 
and        log  7  +  colog  5  -  10  =  0.8451  +  9.3010  — 10, 

=  0.1461, 
it  is  evident  that, 

The  addition  of  a  cologarithm  —  10  is  equivalent  to  the 
subtraction  of  a  logarithm. 

The  steps  that  lead  to  this  result  are : 

?  =  7x  ^, 
therefore,  log  ?  =  log  (7  x  • )  =  log  7  +  log  \.  §  304. 

But  log  \  =  colog  5—10.  §  309. 

Hence,  log  |  =  log  7  +  colog  6  —  10. 

Therefore, 

312.  The  logarithm  of  a  quotient  tnay  be  found,  by  add- 
ing together  the  logarithm  of  the  dividend  and  the  cologa- 
rithm of  the  divisor,  and  stibtracting  10  from  the  result. 

In  finding  a  cologarithm  when  the  characteristic  of  the  logarithm 
is  a  negative  number,  it  must  be  observed  that  the  subtraction  of  a 
negative  number  is  equivalent  to  the  addition  of  an  equal  positive 
number. 

Thus,  log  ^-^  =  log  5  +  colog  0.002  -  10, 

=  0.6990  +  12.6990  -  10, 
=  3.3980. 

Here  log.  0.002  =  3.3010,  and  in  subtracting  —  3  from  9  the  result 

is  the  same  as  adding  +  3  to  9. 

2 
Again,  log  —  =  log  2  +  colog  0.07  -  10, 

=  0.3010  +  11.1549-10, 
z=  1.4559, 


268  ALGEBRA. 

Also,       log  ^  =  8.8451  -  10  +  9.0970  -  10, 

=  17.9421  -  20, 

=  7.9421  -  10. 
Here,  log  2^  =  3  log  2  =  3  X  0.3010  =  0.9030. 

Hence,  colog  2^  =  10  —  0.9030  =  9.0970. 


Exercise  111. 

Given:  log  2  =  0.3010;  log  3  =  0.4771;  log  5  =  0.6990; 
log  7  =  0.8451. 

Find  logarithms  for  the  following  quotients  : 


^•1- 

7. 

6, 
3* 

13. 

0.05 
3 

19. 

0.05 
0.003 

25. 

0.022 
33 

M- 

8. 

5 
2* 

14. 

0.005 
2 

20. 

0.007 
0.02 

26. 

33 
0.022 

-I- 

9. 

7 
3* 

15. 

0.07 
5 

21. 

0.02 
0.007 

27. 

78 

0.022 

-1- 

10. 

7^ 
2" 

16. 

5 
0.07 

22. 

0.005 
0.07 

28. 

0.073 
0.0033 

-f- 

11. 

3 

2' 

17. 

3 
0.007 

23. 

0.03 

7 

29. 

0.0052 
73 

-I- 

12. 

7 
0.5' 

18. 

0.003 

7 

24. 

0.0007 
0.2 

30. 

73 
0.0052 

313.  A  table  oi  four-place  logarithms  is  here  given,  which 
contains  logarithms  of  all  numbers  under  1000,  the  decimal 
point  and  characteristic  being  omitted.  The  logarithms  of 
single  digits  1,  8,  etc.,  will  be  found  at  10,  80,  etc. 

Tables  containing  logarithms  of  more  places  can  be  pro- 
cured, but  this  table  will  serve  for  many  practical  uses, 
and  will  enable  the  student  to  use  tables  of  six-place, 
seven-place,  and  ten-place  logarithms  in  work  that  requires 
greater  accuracy. 


LOGARITHMS.  269 

314.  In  working  with  a  four-place  table,  the  numbers 
corresponding  to  the  logarithms,  that  is,  the  antilogarithms, 
as  they  are  called,  may  be  carried  to  four  significant  digits. 

To  Find  the  Logarithm  of  a  Number  in  this  Table. 

315.  Suppose  it  is  required  to  find  the  logarithm  of 
65.7.  In  the  column  headed  "  K  "  look  for  the  first  two 
significant  figures,  and  at  the  top  of  the  table  for  the  third 
significant  figure.  In  the  line  with  0)5,  and  in  the  column 
headed  7,  is  seen  8176.  To  this  number  prefix  the  charac- 
teristic and  insert  the  decimal  point.     Thus, 

log  65.7  =  1.8176. 

Suppose  it  is  required  to  find  the  logarithm  of  20347. 
In  the  line  with  20,  and  in  the  column  headed  3,  is  seen 
3075;  also  in  the  line  with  20,  and  in  the  4  column,  is 
seen  3096,  and  the  difference  between  these  two  is  21. 
The  difference  between  20300  and  20400  is  100,  and  the 
difference  between  20300  and  20347  is  47.  Hence,  ^^^  o^ 
21  =  10,  nearly,  must  be  added  to  3075.  That  is, 
log  20347  =  4.3085. 

Suppose  it  is  required  to  find  the  logarithm  of  0.0005076. 
In  the  line  with  50,  and  in  the  7  column,  is  seen  7050 ;  in 
the  8  column,  7059 :  the  difference  is  9.  The  difference 
between  5070  and  5080  is  10,  and  the  difference  between 
5070  and  5076  is  6.  Hence,  y%  of  9  =  5  must  be  added  to 
7050.     That  is,     ^^^  0.0005076  =  6.7055  - 10. 

To  Find  a  Number  when  its  Logarithm  is  Given. 

316.  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  1.9736. 

Look  for  9736  in  the  table.  In  the  column  headed  "  N," 
and  in  the  line  with  9736,  is  seen  94,  and  at  the  head  of 


270 


ALGEBRA. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2830 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2072 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26  ' 

"4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

6211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5599 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

0274 

6284 

6294' 

6304 

6314 

6325 

43 

6335 

5345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875' 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7l01 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

LOGARITHMS. 


271 


N 

0 

1   2 

3 

4 

6 

6 

7 

8 

9 

55 

7401  7412 

7410 

7427 

7435 

7443 

7451 

7450 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7651 

57 

7559 

7566 

7574 

7582 

7580 

7507 

7604 

7612 

7610 

7627 

58 

7634 

7642 

7640 

7657 

7664 

7672 

7TT7^ 

7686 

7604 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7830 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7880 

7806 

7003 

7010 

7017 

62 

7924 

7931 

7938y 

7045 

7052 

7050 

7066 

7073 

7080 

7087 

63 

7993 

8000 

8007 

8014 

8021 

8028 

B035 

8041 

8048 

8065 

64 

8062 

8060 

8075 

8082 

8080 

8006 

8102 

8100 

8116 

8122 

65 

8129 

8136 

8142 

8140 

8156 

8162 

8160 

8176 

8182 

8180 

66 

8195 

8202 

8200 

8215 

8222 

8228 

8235 

8241 

8248- 

8264 

67 

8261 

8267 

8274 

8280 

8287 

8203 

8200 

8306 

8312 

8310 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8305 

8401 

8407 

8414 

8420 

8426 

8432 

8430 

8446 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8438 

8404 

8500 

8606 

71 

72 

8513 

8573 

8510 
8570" 

8525 

Mm 

8531 
8501 

8537 
'8597 

8543 
8603 

8549 

8555 
8615 

^561 

8021 

8567 
8627 

8609 

73 

8633 

8639 

8646 

8651 

8667 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8608 

8704 

8710 

8716 

8722 

8727 

8733 

8730  8745  1 

75 

8751 

8756 

8762 

8768 

8774 

8770 

8785 

8701 

8707 

8802 

76 

77 

8808 
8865 

8814 
8871 

8820 
8876 

8825 

8831 

88.^7 

8842 

8848 

8854 
^010 

8859 
8916 

8882 

8887 

8803 

8899 

8004 

78 

8021 

8027 

8032 

8038 

8043 

8040 

8954 

8060 

8065 

8971 

79 

8976 

8082 

8087 

8003 

8008 

0004 

9000 

0015 

0020 

9025 

80 

9031 

9036 

9042 

0047 

0053 

0058 

0063 

0069 

0074 

90X9 

81 

Q085 

0000 

9006 

0101 

0106 

0112 

0117 

9122 

0128 

9133 

82 

oIW" 

0143 

0140 

0154 

0150 

0165 

0170 

9175 

0180 

9186 

83 

9101 

0106 

0201 

0206 

9212 

0217 

0222 

9227 

0232 

9238 

84 

9243 

0248 

9253' 

9258 

0263 

0260 

0274 

9279 

0284 

9289 

85 

9204 

0200 

9304 

9309 

0315 

0320 

0325 

9330 

0336 

9340 

86 

9345 

0350 

9355 

9360 

mm 

0370 

0375 

0380 

0385 

9390 

87 

0395 

0400 

0405 

9410 

0415 

0420 

9425 

0430 

0435 

0440 

88 

9445 

0450 

0455 

0460 

0465 

0460 

9474 

0470 

0484 

0480 

89 

9494 

0400 

0504 

0500 

0513 

0518 

9523 

0528 

0533 

0538 

90 

9542 

0547 

0552 

0557 

0562 

0566 

9571 

0576 

0581 

0586 

91 

0590 

0505 

0600 

0605 

0600 

0614 

9610 

0624 

0628 

0633 

92 

9638 

0643 

0647 

0652 

0657 

0661 

0666 

0671 

0675 

0680 

93 

0685 

0680 

0604 

0600 

9703 

0708 

0713 

0717 

0722 

0727 

94 

9731 

0736 

0741 

0745 

9750 

0754 

0750 

0703 

9768 

0773 

95 

9777 

0782 

0786 

0701 

0705 

0800 

0805  0800 

9814 

0818 

96 

9823 

0827 

0832 

0836 

0841 

0845 

0850  0854 

0859 

0863 

97 

9868 

0872 

9877 

0881 

0886 

0800 

0804  0800. 

9903 

0008 

98 

0012 

0017 

0021 

0026 

0030 

9034 

0030  0043 

0048 

0052 

99 

0050 

0061 

0065 

0060  0074 

9078  0083  1  0087 

0001 

0006 

272  ALGEBRA. 

the  column  in  which  9736  stands  is  seen  1.  Therefore, 
write  941,  and  insert  the  decimal  point  as  the  character- 
istic directs.     That  is,  the  number  required  is  94.1. 

Suppose  it  is  required  to  find  the  number  of  which  the 
logarithm  is  3.7936. 

Look  for  7936  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to 
be  7931  and  7938 ;  their  difference  is  7,  and  the  difference 
between  7931  and  7936  is  5.  Therefore,  |  of  the  differ- 
ence between  the  numbers  corresponding  to  the  mantissas, 
7931  and  7938,  must  be  added  to  the  number  corresponding 
to  the  mantissa  7931. 

The  number  corresponding  to  the  mantissa  7938  is  6220. 
The  number  corresponding  to  the  mantissa  7931  is  6210. 
The  difference  between  these  numbers  is  10, 
and  6210  + 1  of  10  =  6217. 

Therefore,  the  number  required  is  6217. 

Suppose  it  is  required  to  find  the  number  of  which  the 
logarithm  is  7.3882  —  10. 

Look  for  3882  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to 
be  3874  and  3892  ;  their  difference  is  18,  and  the  difference 
between  3874  and  3882  is  8.  Therefore,  ^%  of  the  differ- 
ence between  the  numbers  corresponding  to  the  mantissas, 
3874  and  3892,  must  be  added  to  the  number  corresponding 
to  the  mantissa  3874. 

The  number  corresponding  to  the  mantissa  3892  is  2450. 
The  number  corresponding  to  the  mantissa  3874  is  2440. 
The  difference  between  these  numbers  is  10, 
and  2440  -f  |  of  10  =  2444. 

Therefore,  the  number  required  is  0.002444. 


^ 


LOGARITHMS.  \  273 


Exercise  112. 
Find  logarithms  of  the  following  numbers  ; 

1.  60.  6.    3780.  11.    70633.  16.  877.08. 

2.  101.  7.    54327.         12.    12028.  17.  73.896. 

3.  999.r         8.    90801.^      13.   0.00987.7  18.  7.0699.1^ 

4.  9901.i        9.    10001.5^     14.    0.87701.p  19.  0.0897// 

5.  5406.  5     10.    10010.4     15.    1.0001.^  20.  99.778./:^ 

Find  antilogarithms  to  the  following  logarithms  : 

21.  4.2488.  25.    4.7317.  29.    9.0410  —  10. 

22.  3.6330.  26.    1.9730.  ^0.    9.8420  —  10. 

23.  2.5310.  /  3  27.    9.8800  —  10./^  31.    7.0216  — 10. /■> 

24.  1.9484.    //^         28.    0.2787.       I'i         32.    8.6580  —  10.   /p" 

Ex.    Find  the  product  of  908.4  X  0.05392  X  2.117. 

log     908.4  =  2.9583 

log  0.06392  =  8.7318 -10 

log      2.117=0.3257 

2.0158  =  log  10.3.7.   Ans. 

Find  by  logarithms  the  following  products  : 

33.  948.76  X  0.043875.  J>  35.   830.75  X  0.0003769. 

34.  3.4097  X  0.0087634.  »A   36.   8.4395  X  0.98274. 


317.  When  any  of  the  factors  are  negative,  find  their 
logarithms  without  regard  to  the  signs  ;  write  the  letter  n 
after  the  logarithm  that  corresponds  to  a  negative  number. 
If  the  number  of  logarithms  so  marked  is  odd,  ttie  product 
is  negative;  if  even,  the  product  is  positive. 


274 


ALGEBRA. 


'S 


Find  the  products  of : 


7 


37.  7564  X(- 0.003764).    f    39.  —5.840359  X  (-0.00178). 
y38.  3.7648  X(- 0.083497).  S  40.  -8945.07X73.846. 

8.3709  X  834.637 


Ex.    Find  the  quotient  of 


7308.946 


log  8.3709  =  0.9227 

log        834.637  =  2.9215 
colog  7308.946  =  6.1362-10 

9.9804  -  10  =  log  0.9558.    Arts. 


Find  the  quotients  of 

70654 
41. 


S"'    54013 
N"^  43. 


\  44. 
\    46. 


93078 

-^.32165 
0.07891' 

650^ 
90761' 

7.652 
—  0.06875' 


46. 


47. 


48. 


49. 


50. 


0.07654 


83.947  X  0.8395 

7564  X  0.07643 
809i5  X  0.09817' 

89  X  753  X  0.0097 
36709X0.08497 

413  X  8.17  X  3182 
915  X  728  X  2.315* 


ir 


212  X(- 6.12)  X(- 2008) 
365  X  (-531)  X  2.576 


^ 


Ex.    Find  the  cube  of  0.0497. 


log  0.0497  =  8.6964  -  10 
3 

}1228. 

6.0892  —  10  =  log  0.00( 

Ans. 

Find  by  logarithms  : 
\\l.    6.05^.          55.'^0.78765«.     59.    (lOf)^ 

63. 

m\t"- 

(r.^^52.    1,051".        56.    0.691"J>^\    60.    (1^)«. 

64. 

(lA/-^-, 

^\.53.    1.1768^       57.    (^f)".           61.    (|fi)«. 

65. 

(8|)H  '■ 

ov54.    1.3178^".     58.    {\\y,            62.    (7^^)"-^ 

66. 

(5§})»» 

LOGARITHMS. 

ZiiJ 

Ex.    Find  the  fourth  r^ 

oot  of  0.00862. 

log  0.00862  = 

:    7.9355-10 
30.          -  30 

4)37.9355  -  40 

9.4839 -10  =  log  0.3047. 

Ans. 

Find  by  logarithms : 
67.    7i            70.    8'379A. 

73.    0.17643i. 

76. 

u^^y- 

68.    Hi         71.    906.80i. 

74.    2.5637A. 

77. 

(my- 

69.    783i.       72.    8.1904i 

75.    (4e)i. 

78. 

(ii?t)»- 

Find  by  logarithms  the  values  of  : 

0.0075433^  X  78.343  X  8172.4i  X  0.00052 


V) 


64285*  X  154.27*  X  0.001  X  586. 79J 
15.832^  X  5793.6i  X  0.78426 


^  \0.0003274X  768.94=^  X  3015.3  X0.007i 

5/7.1895  X  4764.2^  X  0.00326^ 
^^'    \0.00048953  X  457«  X  5764.4=^* 

/3.1416  X  4771.21  X  2.7183* 


82. 


4 


'^  \30.103*X  0.4343^X69.897* 

W0.03271*'*  X  53.429  X  0.77542« 


83. 


84. 


85. 


\  32.769  X  0.000371* 

732.056^  X  0.0003572*  X  89793 


42.2798«  X  3.4574  X  0.0026518* 

7932  X  0.00657  X  0.80464 
0.03274  X  0.6428 


3/   7.1206  X  V0.13274  X  0.057389 
86.    yj 


87. 


VO.43468  X  17.385  X  VO.0096372 
3.075526^  X  5771.2*  X  0.0036984*  X  7.74  ^  « 


72258X327.93^X86.97* 


}■ 


276  ALGEBRA. 

318.  Since  any  positive  number  other  than  1  may  be 
taken  as  the  base  of  a  system  of  logarithms,  the  following 
general  proofs  to  the  base  a  should  be  noticed. 

I.  The  logarithm  of  the  prodicct  of  two  or  more  numbers  is 
equal  to  the  sum  of  the  logarithms  of  the  numbers. 

For,  let  m  and  n  "be  two  numbers,  and  x  and  y  their  logarithms. 
Then,  by  the  definition  of  a  logarithm,  m  =  a^,  and  n  =  av. 
Hence,  mX  n  =  a''X  ay  =  a^'+y. 

.-.  log  {mx  n)  =  x-{-y, 

=  log  m  +  log  n. 

In  like  manner,  the  proposition  may  be  extended  to  any  number 
of  factors. 

II.  The  logarithm  of  a  quotient  is  equal  to  the  logarithm 
of  the  dividend  minus  the  logarithm  of  the  divisor. 

For,  let  m  and  n  be  two  numbers,  and  x  and  y  their  logarithms. 
Then  m  =  a^,  and  n  =  av. 

Hence,  m-^n^a^-r-  av  =  a^-v. 

.:  log  {m-7-n)  =  x  —  y, 

=  log  m  —  log  n. 

From  this  it  follows  that  log  —  =  log  1  —  log  m. 
m 

But,  since  log  1  =  0,  log  —  =  —  log  m. 


III.  The  logarithm,  of  a  power  of  a  number  is  equal  to 
the  logarithm  of  the  number  multiplied  by  the  exponent  of 
the  power. 

For,  let  X  be  the  logarithm  of  m. 
Then  m  =  a^, 

and  mp  =  {a^)p  =  a^*. 

.'.  log  7nP  =  px, 

=  p  log  m. 


LOGARITHMS.  277 

IV.  The  logarithm  of  the  root  of  a  number  is  equal  to  the 
logarithm,  of  the  number  divided  by  the  index  of  the  root. 

For,  let  X  be  the  logarithm  of  m. 
Then  m  =  a^, 

]  1  X 

and  m'-  =  (a^)^  =  a^ 

1      X      losr  m 

.'.  log  mr  =  -  =  —^ 

^  r  r 

319.  An  exponential  equation,  that  is,  an  equation  in 
which  the  exponent  is  the  unknown  quantity,  is  easily 
solved  by  logarithms.    • 

For,  let  a^  =  m. 

Then  log  a^  =  log  m. 

.'.  X  log  a  =  log  m. 

_  log  m 

~  log  a 

Ex.   Find  the  value  of  x  in  81'  =  10. 

8P  =  10, 
^  log  10 
^     log  81* 
.••  log  X  =  log  log  10  +  colog  log  81, 
=  0  +  9.7193  -  10. 
.-.  a;  =  0.624. 

320.  Logarithms  of  numbers  to  any  base  a  may  be  con- 
verted into  logarithms  to  any  other  base  b  by  dividing  the 
computed  logarithms  by  the  logarithm  of  b  to  the  base  a. 

For,  let  log  m  =  y  to  the  base  6, 

and  log  6  =  X  to  the  base  a. 

Then  m  =  b'J,  and  b  =  a^. 

.:  m  =  (a^)J'  =  a'^y. 

.:  log  m  (to  base  a)=xy  =  log  b  (to  base  a)  X  log  m  (to  base  b). 

1  ,^    -u       ^.^      log  m  (to  base  a) 

.:  log  m  (to  base  b)  =  ,  °      ,;    , 1 

^      ^  '      log  6  (to  base  a) 

This  is  usually  written,  logs  m  =  - — ^* 


CHAPTEE   XX. 

E-ATio,  Proportion,  and  Variation. 

321.  The  relative  magnitude  of  two  numbers  is  called 
their  ratio,  when  expressed  by  the  fraction  which  the  first 
is  of  the  second.  , 

Thus,  the  ratio  of  6  to  3  is  indicated  by  the  fraction  |,  which  is 
sometimes  written  6  : 3. 

322.  The  first  term  of  a  ratio  is  called  the  antecedent, 
and  the  second  term  the  consequent.  When  the  antecedent 
is  equal  to  the  consequent,  the  ratio  is  called  a  ratio  of 
equality ;  when  the  antecedent  is  greater  than  the  conse- 
quent, the  ratio  is  called  a  ratio  of  greater  inequality ;  when 
less,  a  ratio  of  less  inequality. 

323.  When  the  antecedent  and  consequent  are  inter- 
changed, the  resulting  ratio  is  called  the  inverse  of  the 
given  ratio. 

Thus,  the  ratio  3  :  6  is  the  inverse  of  the  ratio  6:3. 

324.  The  ratio  of  two  quantities  that  can  be  expressed 
in  integers  in  terms  of  a  common  unit  is  equal  to  the  ratio 
of  the  two  numbers  by  which  they  are  expressed. 

Thus,  the  ratio  of  $9  to  $11  is  equal  to  the  ratio  of  9 :  11 ;  and  the 
ratio  of  a  line  2|-  inches  long  to  a  line  3f  inches  long,  when  both  are 
expressed  in  terms  of  a  unit  -^^  of  an  inch  long,  is  equal  to  the  ratio 
of  32  to  45, 

325.  Two  quantities  different  in  kind  can  have  no  ratio, 
for  then  one  cannot  be  a  fraction  of  the  other. 


RATIO.  279 

326.  Two  quantities  that  can  be  expressed  in  integers 
in  terms  of  a  common  unit  are  said  to  be  coinmensurable. 
The  common  unit  is  called  a  comvion  measure,  and  each 
quantity  is  called  a  multiple  of  this  common  measure. 

Thus,  a  common  measure  of  2^  feet  and  3f  feet  is  ^  of  a  foot,  which 
is  contained  15  times  in  2^  feet,  and  22  times  in  3|-  feet.  Hence,  2^ 
feet  and  3|-  feet  are  multiples  of  ^  of  a  foot,  2|  feet  being  obtained 
by  taking  ^  of  a  foot  15  times,  and  8f  by  taking  |  of  a  foot  22  times. 

327.  When  two  quantities  are  incommensurable,  that  is, 
have  no  common  unit  in  terms  of  which  both  quantities 
can  be  expressed  in  integers,  it  is  impossible  to  find  a  frac- 
tion that  will  indicate  the  exact  value  of  the  ratio  of  the 
given  quantities.  It  is  possible,  however,  by  taking  the 
unit  sufficiently  small,  to  find  a  fraction  that  shall  differ 
from  the  true  value  of  the  ratio  by  as  little  as  we  please. 

Thus,  if  a  and  h  denote  the  diagonal  and  side  of  a  square, 

Now\^=  1.41421356 ,  a  value  greater  than  1.414213,  but  less 

than  1.414214. 

If,  then,  a  millionth  part  of  h  be  taken  as  the  unit,  the  value  of  the 
ratio  ^  lies  between  mtf^f  and  rUMM»  ^"^^  therefore  differs  from 
either  of  these  fractions  by  less  than  t^^^ij^^^. 

By  carrying  the  decimal  farther,  a  fraction  may  be  found  that  will 
differ  from  the  true  value  of  the  ratio  by  less  than  a  billionth^  tril- 
lionth,  or  any  other  assigned  value  whatever. 

328.  Expressed  generally,  when  a  and  b  are  incommen- 
surable, and  b  is  divided  into  any  integral  number  (n)  of 
equal  parts,  if  one  of  these  parts  be  contained  in  a  more 
than  m  times,  but  less  than  m-\-l  times,  then 

am,  m-{-l 

7  >  — J  but  <  —  : 

b        n  n 

that  is,  the  value  of  -  lies  between  —  and ^• 

b  n  n 


280  ALGEBKA. 

The  error,  therefore,  in  taking  either  of  these  values  for 

-  is  <  —     But  by  increasing  n  indefinitely,  -  can  be  made 

to  decrease  indeiinitely,  and  to  become  less  than  any  as- 
signed value,  however  small,  though  it  cannot  be  made 
absolutely  equal  to  zero. 

329.  The  ratio  between  two  incommensurable  quantities 
is  called  an  incommensurable  ratio. 

330.  As  the  treatment  of  Proportion  in  Algebra  depends 
upon  the  assumption  that  it  is  possible  to  find  fractions 
which  will  represent  the  ratios,  and  as  it  appears  that  no 
fraction  can  be  found  to  represent  the  exact  value  of  an 
incommensurable  ratio,  it  is  necessary  to  show  that  two 
incommensurable  ratios  are  equal  if  their  true  values  always 
lie  between  the  same  limits,  however  little  these  limits  differ 
from  each  other. 

Let  a  :  h  and  c  :  c?  be  two  incommensurable  ratios. 

Suppose  the  true  values  of  the  ratios  a  :  h  and  c:d  lie  between 

7Yl  7Th  "4"  1 

—  and .     Then  the  difference  between  the  true  values  of  these 

n  n  ^  1 

ratios  is  less  than  - ,  however  small  the  value  of  -  may  be.  §  328. 

1      ^  ^1 

But  since  -  can  be  made  to  approach  zero  at  pleasure,  -  can  be 

made  less  than  any  assumed  difference  between  the  ratios. 

Therefore,  to  assume  any  difference  between  the  ratios  is  to  assume 

it  possible  to  find  a  quantity  that  for  the  same  value  of  -  shall  be  both 

greater  and  less  than  -  ;  which  is  a  manifest  absurdity. 

Hence,  a:h=  c-.d. 

331.  It  will  be  well  to  notice  that  the  word  limit  means 
a  fixed  value  from  which  another  and  variable  value  may 
be  made  to  differ  by  as  little  as  we  please  ;  it  being  impos- 
sible, however,  for  the  difference  between  the  variable 
value  and  the  limit  to  become  absolutely  zero. 


RATIO.  281 

332.  A  ratio  will  not  be  altered  if  both  its  terms  be  multi- 
plied by  the  same  number. 

For  the  ratio  a :  6  is  represented  by  - ,  the  ratio  ma:mb  is  repre- 

^   -  ,     ma         -    .        ma      a 
sented  by  —r ;  and  since  ^  =  t  ^  .'.  ma:mb  =  a:b. 
mb  mb      b 

333.  A  ratio  will  be  altered  if  different  multipliers  of  its 
terms  be  taken;  and  will  be  increased  or  diminished  accord- 
ing as  the  multiplier  of  the  antecedent  is  greater  or  less  than 
that  of  the  consequent. 

For,     ma  :  nb  will  be  >  or  <  a  :  6 

,.  ma  .   ^  ^  a  /      na\ 

a«cordingas  — ^>or<-(=-) 

as  ma  is  >>  or  <;  na, 

as  7/1    is  >•  or  <;  n. 

334.  A  ratio  of  greater  inequality  will  be  diminished^ 
and  a  ratio  of  less  inequality  increased  by  adding  the  same 
number  to  both  its  terms. 

For,     a-Y  x:b-\-  X  is>or-<a:6 

,.  a  +  X  .   ^         ^  a 

according  as       z—. —  is  >  or  <  7 , 
b  +  X  b 

as    ab  +  6x  is  >  or  <Cah  +  ax, 

as  6x  is  >•  or  •<  oa;, 

as  6  is  >  or  <  a. 

335.  A  ratio  of  greater  inequality  ivlll  be  increased,  and 
a  ratio  of  less  inequality  diiiiinished,  by  subtracting  the  same 
number  from,  both  its  terms. 

For,  a— X: 6— X  will  be >  or  <  a:6 

,.  a  —  X  .   ^         ^a 

according  as is  >  or  <  r » 

b  —  X     ^  b 

as    ab  —  6x  is  >  or  <  a6  —  ox, 

as  ox  is  >►  or  -<  6x, 

as  a  is  >>  or  <;  6. 


282  ALGEBRA. 

336.  Ratios  are  compounded  by  taking  the  product  of 
the  fractions  that  represent  them. 

Thus,  the  ratio  compounded  of  a :  6  and  c  :  d  is  found  by  taking  the 

product  of  -r  and  -  =  73  * 
^  b         d      bd 

The  ratio  compounded  of  a  :  &  and  a  :  6  is  the  duplicate  ratio  a^ :  6^, 

and  the  ratio  compounded  of  a  :  6,  a  :  6,  and  a  :  6  is  the  triplicate  ratio 

337.  Ratios  are  compared  by  comparing  the  fractions 
that  represent  them. 


Thus, 

a  :  6  is  >  or  <  c  :  1 

according  as 

a  .   ^          ^  c 
^is>or<-- 

as 

ad  .   ^          ^bc 

aa 

ad  is  >  or  <  be. 

■ 

Exercise  113. 

1.  Write  down  the  ratio  compounded  of  3 : 5  and  8 : 7. 

Which  of   these  ratios   is  increased,  and  which  is 
diminished  by  the  composition? 

2.  Compound  the  duplicate  ratio  of  4 :  15  with  the  tripli- 

cate of  5 : 2. 

3.  Show  that  a  duplicate  ratio  is  greater  or  less  than  its 

simple  ratio  according  as  it  is  a  ratio  of  greater  or 
less  inequality. 

4.  Arrange  in  order  of  magnitude  the  ratios  3:4;  23:25; 

10  :  11 ;  and  15  :  16. 

5.  Arrange  in  order  of  magnitude 

a-\-b:a  —  b  and  a^-{-b^:  a^  —  b^,  if  a'>b. 

Pind  the  ratios  compounded  of : 

6.  3  :  5  ;  10  :  21 ;  14  :  15.  7.7:9;  102  :  105  ;  15  :  17. 


RATIO.  283 

a^4-ax4-a^  ,  a^  —  ax-{-x^ 

S-   1 2 — I -> ^and- i 

a"*  —  a'^x  -f-  ax-  —  x"^  a-f  x 

x^  —  9x-i-20       ^x^  —  13x-i-4.2 

9.   7, — 7^ and ^ — 

x'^  —  ox  x^  —  ox 

10.  a^h'.a-h',  a?-\-l?'.{a-\-}jf',   {a? -hy '.  a^-h\ 

11.  Two  numbers  are  in  the  ratio  2  :  3,  and  if  9  is  added  to 

each,  they  are  in  the  ratio  3  :  4.     Find  the  numbers. 

(Let  2x  and  3x  represent  the  numbers. ) 

12.  Show  that  the  ratio  a\h  \^  the  duplicate  of  the  ratio 

a-\-c\h-\-c,  if  c'^=ah. 

13.  Find  two  numbers  in  the  ratio  3  :  4,  of  which  the  sum 

is  to  the  sum  of  their  squares  in  the  ratio  of  7  to  50. 

14.  If  five  gold  coins  and  four  silver  ones  are  worth  as  much 

as  three  gold  coins  and  twelve  silver  ones,  find  the 
ratio  of  the  value  of  a  gold  coin  to  that  of  a  silver 
one. 

15.  If  eight  gold  and  nine  silver  coins  are  worth  as  much 

as  six  gold  and  nineteen  silver  coins,  find  the  ratio 
of  the  value  of  a  silver  coin  to  that  of  a  gold  one. 

16.  There  are  two  roads  from  A  to  B,  one  of  them  14 

miles  longer  than  the  other ;  and  two  roads  from 
B  to  C,  one  of  them  8  miles  longer  than  the  other. 
The  distance  from  A  to  B  is  to  the  distance  from  B 
to  C,  by  the  shorter  roads,  as  1  to  2 ;  by  the  longer 
roads,  as  2  to  3.     Find  the  distances. 

17.  What  must  be  added  to  each  of  the  terms  of  the  ratio 

m :  n,  that  it  may  become  equal  to  the  ratio  j^  -2^ 

18.  A  rectangular  field  contains  5270  acres,  and  its  length 

is  to  its  breadth  in  the  ratio  of  31 :  17.  Find  its 
dimensions. 


284  ALGEBRA. 


Proportion. 

338.  An  equation  consisting  of  two  equal  ratios  is  called 
a  proportion;  and  the  terms  of  the  ratios  are  called  pro- 
portionals. 

339.  The  algebraic  test  of  a  proportion  is  that  the  two 
fractions  which  represent  the  ratios  shall  be  equal. 

(X       c 
Thus,  the  ratio  a :  h  will  be  equal  to  the  ratio  c:d  if  -  =  - ;  and 

the  four  numbers  a,  6,  c,  d  are  called  proportionals,  or  are  said  to  be 
in  proportion. 

340.  If  the  ratios  a :  b  and  c :  d  form  a  proportion,  the 
proportion  is  written       a:b  =  G:d,     and  is  read 

the  ratio  of  a  to  &  is  equal  to  the  ratio  of  c  to  d ', 
or  a:  b  ::  c:  d,     and  is  read 

a  is  to  b  in  the  same  ratio  as  c  is  to  d. 

The  first  and  last  terms,  a  and  d,  are  called  the  extremes. 

The  two  middle  terms,  b  and  c,  are  called  the  means. 

341.  When  four  numbers  are  in  proportion,  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 

For,  if  a  :b  :  :  c  :d, 

^.  a      c 

then  T  =  'i' 

b      d 

By  multiplying  by  bd,  ad  =  be. 

342.  If  the  product  of  two  numbers  is  equal  to  the  product 
of  two  others,  either  two  may  be  made  the  extremes  of  a  pro- 
portion and  the  other  two  the  means. 

For,  if  •  ad  —  be, 

by  dividing  by  6d!,       '    ^  =  5^' 
a      c 

.'.  a:b  :  :e:d. 


PROPORTION.  286 

343.  The  equation  ad^hc  gives 

he  ad 

SO  that  an  extreme  may  be  found  by  dividing  the  product 
of  the  means  by  the  other  extreme ;  and  a  mean  may  be 
found  by  dividing  the  product  of  the  extremes  by  the  other 
mean. 

344.  If  four  quantities,  a,  h,  c,  d,  be  in  proportion,  they 
will  be  in  proportion  by  : 

I.   Inversion. 

That  is,  b  will  be  to  a  as  c?  is  to  c. 


For, 

if 

a  :  6  : :  c  :  d, 

then 

a      c 

and 
or 

b_d^ 

a      c 

.\b:a::d.C. 

345.    IT.    Composition. 

That  is,     a-\-h  will  be  to  Z/  as  c  +  c?  is  to  d. 


For,  if 

a:6::c  :d, 

then 

a      c 

and 

^'=^^' 

or 

a+6      c+d 
b             d 

a  +  b:b'.:c  +  d'.d. 

346.   III.   Division. 

That  is,     a  —  b  will  be  to  ^>  as  c  —  c?  is  to  d. 

For,  if 

a:b::c:d^ 

then 

a       c 

286  ALGEBRA. 

and  7  —  1  =  -  —  1, 

b  d       ' 

a  —  b      c  —  d 

or  —r-  =  — r-  • 

b  d 

.'.a  —  b:b::c  —  d:d. 

347.   IV.    Composition  and  Division. 

That  is,  a~\-b  will  be  to  a  —  b  SiS  c-{-  d  is  to  c  —  d, 

-r,      J-        TT  a  +  b     c  +  d 

For,  from  II.,         — r—  =  — r—  » 
0  a 

and  from  III. , 


By  dividing, 


b  d 

a  +  6      c-\-  d 


a  —  b      c  —  d 
.-.  a  ■}-  b  :  a  —  b  : :  c  -h  d  :  c  —  d. 


348.   When  the  four  quantities  a,  b,  c,  d  are  all  of  the 
same  kind,  they  will  be  in  proportion  by : 


V.   Alternation. 

That  is, 

a  will  be  to  c  as  ^  is 

to  d. 

For,  if                    a:l 

)  : :  C 

:d, 

then 

b~ 

'  d 

By  multiplying  by  -? 

ab 
be 

be 

'■7d' 

or 

a  _ 
c 

.6, 
'  d 

.-.  a 

:  c  :: 

bid. 

349.   From  the  proportion  aicr.bid   may  be  obtained 
by: 

VI.    Composition.     a-{-c:c::b-{-d:d. 

VII.    Division.  a  —  c:c::b  —  d:d. 

VIII.    Composition  and  Division.  a-\-c:a—c::b-\-d:d—d. 


PROPORTION.  287 

350.  In  a  series  of  equal  ratios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

„       .-  a      c       e      g 

For,  if  -  =  -  =  —  =  |, 

b      d      f      h 

r  may  be  put  for  each  of  these  ratios. 

Then    -  =  r,  ^  =  r,  -  =  r,  ^=r. 

.*.  a=br,  c=  dr,  e=  fr,  g=  hr. 
,'.a  +  c  +  e  +  g={b-{-d-\-f+h)r. 

a+  c  +  e  +  flf  _  _a 
'''b  +  d+f  +  h~^~b' 
.-.  a  +  c  ■{-  e  +  g  :  b  -\-  d  -{•  f  +  h  ::  a :  b. 

In  like  manner  it  may  be  shown  that 

ma-{-  nc-\-pe-^  qg  :  mb-\-  nd  +  pf+  qh::  a:b. 

351.  If  a,  b,  c,  d  be  in  continued  proportion,  that  is,  if 
a:b  =  b  :  c  =  c:  d,  then  will  a  :  c=^a^  :b^  and  a:d  =  a^ :  b^. 


For, 
Hence, 
or 

So 


a 
b~ 

b_ 
c 

c 

d 

^ 

,b_ 

'  c 
a_ 
c~ 

V 

.-.  a 

:  C  =: 

:a2: 

62. 

a 
b 

c 

c 

d~ 

a 
d~ 

1x 

68* 

-.X 

a 
6 

.-.  a 

'.d  = 

:a3: 

63. 

352.  If  a,  b,  c  are  proportionals,  so  that  a  :  b  :  :  b  :  c,  then 
b  is  called  a  wean  proportional  between  a  and  c,  and  c  is 
called  a  third  proportional  to  a  and  b. 


288  ALGEBRA. 


If  a  :h:  lb:  c,  then  h  = 

•\lac. 

For,  if 

a:b::b:c, 

then 

a      b 

and 

b'^  =  ac. 

.'.  b  =  Vac, 

353.    The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 

For,  if  a:b::c:d, 

e:f::g:h, 

and  k:l::m:n, 

^,  a      c    e      g    k      m 

then  7  =  3'  7=  I'  1  =  — 

b      d    f     h     I      n 

Hence,  by  finding  the  product  of  the  left  members,  and  also  of  the 
right  members  of  these  equations, 

aek  _  cgm 

bfl  ~  dhn 

.-.  aek  :bfl::  cgm  :  dhn. 

364.   Like  powers,  or  like  roots,  of  the  terms  of  a  propor- 
tion are  in  proportion. 

For,  if  a:b  ::c  :d, 

*u  a      c 

then  7  =  :;• 

b      d 

By  raising  both  sides  to  the  nth  power. 

6«  ~  d« 
.-.  a»:6«::c»:d«. 

By  extracting  the  nth  root, 
1        1 

1  ""    1* 

1111 
.'.  a":6«::  C":d». 


PROPORTION.  289 

355.  If  two  quantities  he  increased  or  diminished  by  like 
parts  of  each,  the  results  will  he  in  the  same  ratio  as  the 
quantities  themselves. 


For, 


that  is, 


a  ±  —  a 
a  n 


n 

.:  a  :  0  ::  a  ±.  —  a  :  0  ±  —  0. 
n  n 

356.  The  laws  that  have  been  established  for  ratios 
should  be  remembered  when  ratios  are  expressed  in  their 
fractional  form. 

,  X    o  -,  x'-^x+l      x'-x-\-2 

(1)  Solve:  ^^-^^  =  ^^^. 

2  x^  2  x* 

By  5347  ¥(^TT)  =  -2(x-2)' 

and  this  equation  is  satisfied,  when  x  =  0  j 

A'    'A-        V.     2X2  1  1 

or,  dividing  by—.  ^+1  =  ^^• 

.■.x  =  h 

(2)  li  a  :  b  : :  c  :  d,  show  that 

a^  +  ab:h^  —  ab::c'-i-cd:d^  —  cd. 

b      d 

then  7  = '  §  347. 

a—o      c  —  d 


and 


that  is. 


a   _    c 
-b~  -d 

_«.v^=-^x^-±^;  §353. 

—  6      a  — &      —  d      c  —  d 

a2  +  a6      c2  +  cd 


62  -  ab      d2  -  cd 
or  a2+a6:62-a6::c2+cd:d2-cd. 


290  ALGEBRA. 

(3)    When  a:h'.:c:d,  and  a  is  the  greatest  term,  show  that 
a  -f-  c?  is  greater  than  h-\-c. 


§346. 


Since 

-=-.  and  a>c, 

then 

Also,  since 

6>d. 
a  —  b      c  —  d 
b     =     d    ' 

and 

then 
By  adding 

b>d, 
a  —  b^c  —  d. 
to  this,    b-\-d=b  +  d, 
a-^d-:>b  +  c. 

Exercise  114. 
If  a:b:  :G:d,  prove  that : 

1.  ma  \7ib::  mc  :  7icl.  4.    a^:b^  :  :c^:  d\ 

2.  3ai-b:h::Sc-{-d:d.  5.    a  :  a-\-b  :  :  c:  c -\- d. 

3.  a-\-2b:b::c-\-2d:d.  6.    a:  a  —  b::c'.G  —  d. 

7.  via-\-nb  :  ma  —  nb  : :  vie  -\-  nd :  mc  —  nd. 

8.  2a-[-3b:3a  —  lb:'.2c-\-M:ZG  —  4.d. 

9.  ma^  -\-  nc^ :  7nb'^ -{-nd'^ :  •.a'^:b^. 

10.    ma'^ -\-  nab  -{-pb^ :  mc^  -\-  ncd  -\-pd'^  wb^'.d^. 

li  a:b::b:c,  prove  that : 
11.    a-\-b  :b-\-c::  a:b.  12.    a^ -\- ab '.  b"^ -{- be '. :  a  :  Cr 

13.    a:c::{a-^by:(b^c)\ 

14.  When  a,  b,  and  c  are  proportionals,  and  a  the  greatest, 

show  that  a-\-c>2b. 

15.  If  — Y^  =  ^  = ,  and  x,  y,  z  are  unequal,  show 

that  l-\-m-\-n=^0. 


tUopofeTioij.  291 

16.  Find  X  when  ic  +  5  :  2a;  —  3  : :  5cc  + 1 :  3ic  —  3. 

17.  Find  x  when  x-\-a:2x  —  h::^x-{-b:4^x  —  a. 

18.  Find  x  when  Vic  +  V^ :  Vic  —  '\lh::a:h. 

19.  Find  £c  and  y  when  ic  : 27  : :  ?/ :  9,  and  x:21  ::2:x  —  y. 

20.  Find  a;  and  y  when  a;  +  ?/  +  l:cc  +  ^  +  2::6:7,  and 

when  7/  +  2x:y  — 2ic::12ic  +  6y  — 3:6y  — 12a;  — 1. 

21.  Find  a;  when  a;^— 4a;+2:  a;^— 2a;— 1::  a;^— 4a;:a;2— 2a;— 2. 

22.  A  railway  passenger  observes  that  a  train  passes  him, 

moving  in  the  opposite  direction,  in  2  seconds ;  but 
moving  in  the  same  direction  with  him,  it  passes  him 
in  30  seconds.     Compare  the  rates  of  the  two  trains. 

23.  A  and  B  trade  with  different  sums.    A  gains  $200  and 

B  loses  $50,  and  now  A's  stock  :  B's  : :  2  :  ^.  But, 
if  A  had  gained  $100  and  B  lost  $85,  their  stocks 
would  have  been  as  15  :  3^-.  Find  the  original  stock 
of  each. 

24.  A  quantity  of  milk  is  increased  by  watering  in  the 

ratio  4  :  5,  and  then  three  gallons  are  sold ;  the 
remainder,  is  mixed  with  three  quarts  of  water,  and 
is  increased  in  the  ratio  6  :  7.  How  many  gallons 
of  milk  were  there  at  first  ? 

25.  In  a  mile  race  between  a  bicycle  and  a  tricycle  their 

rates  were  as  5  : 4.  The  tricycle  had  half  a  minute 
start,  but  was  beaten  by  176  yards.  Find  the  rate 
of  each. 

26.  The  time  which  an  express-train  takes  to  travel  180 

miles  is  to  that  taken  by  an  ordinary  train  as  9  :  14. 
The  ordinary  train  loses  as  much  time  from  stopping 
as  it  would  take  to  travel  30  miles  ;  the  express-train 
loses  only  half  as  much  time  as  the  other  by  stop- 
ping, and  travels  15  miles  an  hour  faster.  What 
are  their  respective  rates  ? 


292  ALGEBRA. 

27.  A  line  is  divided  into  two  parts  in  the  ratio  2:3,  and 

into  two  parts  in  the  ratio  3:4;  the  distance  be- 
tween the  points  of  section  is  2.  Find  the  length 
of  the  line. 

28.  A  railway  consists  of  two  sections ;  the  annual  expendi- 

ture on  one  is  increased  this  year  5^,  and  on  the 
other  4%,  producing  on  the  whole  an  increase  of 
4y%%.  Compare  the  amount  expended  on  the  two 
sections  last  year,  and  also  this  year. 

29.  When  a,  b,  c,  d  are  proportional  and  unequal,  show 

that  no  number  x  can  be  found  such  that  a-\-x, 
b-\-x,  c-\-Xyd-\-x  shall  be  proportionals. 


Variation. 

367.  Two  quantities  may  be  so  related  that,  when  one 
has  its  value  changed,  the  other  will,  in  consequence,  have 
its  value  changed. 

Thus,  the  distance  travelled  in  a  certain  time  will  be  doubled  if  the 
rate  be  doubled.  The  time  required  for  doing  a  certain  quantity  of 
work  will  be  doubled  if  only  half  the  number  of  workmen  be  employed. 

358.  Whenever  it  becomes  necessary  to  express  the 
general  relations  of  certain  kinds  of  quantities  to  each 
other,  without  confining  the  inquiry  to  any  particular  values 
of  these  quantities,  it  will  usually  be  sufficient  to  mention 
two  of  the  terms  of  a  proportion.  In  all  such  cases,  how- 
ever, four  terms  are  always  implied. 

Thus,  if  it  be  said  that  the  weight  of  water  is  proportional  to  its 
volume,  or  varies  as  its  volume,  the  meaning  is,  that  one  gallon  of 
water  is  to  any  number  of  gallons  as  the  weight  of  one  gallon  is  to  the 
weight  of  the  given  number  of  gallons. 


VARIATION.  293 

359.  Quantities  used  in  a  general  sense,  as  distance,  time, 
weight,  volume,  to  which  particular  values  may  be  assigned, 
are  denoted  by  capital  letters.  A,  B,  C,  etc. ;  while  assigned 
values  of  these  quantities  may  be  denoted  by  small  letters, 
a,  h,  c,  etc.  The  letters  A,  B,  C  will  be  understood  to 
represent  ang  numerical  values  that  may  be  assigned  to 
the  quantities ;  and  when  two  such  letters  occur  in  an 
expression  they  will  be  understood  to  represent  any  corre- 
sponding numerical  values  that  may  be  assigned  to  the  two 
quantities. 

360.  When  two  quantities  A  and  B  are  so  connected 
that  their  ratio  is  constatit,  that  is,  remains  the  same  for 
all  corresponding  values  of  A  and  B,  the  one  is  said  to  vary 
as  the  other ;  and  this  relation  is  expressed  hj  A  <^  B  (read 
A  varies  as  B). 

Thus,  the  area  of  a  triangle  with  a  given  base  varies  as  its  altitude  ; 
for,  if  the  altitude  be  changed,  the  area  will  be  changed  in  the  same 
ratio. 

If  this  constant  ratio  be  denoted  by  m,  then  --  =  m,  or 
A  =  mB. 

From  this  equation  ??i  may  be  found  when  two  corre- 
sponding values  of  A  and  B  are  known. 

361.  When  two  quantities  are  so  connected  that  if  one 
be  changed  in  any  ratio,  the  other  will  be  changed  in  the 
inverse  ratio,  the  one  is  said  to  vary  inversely  as  the  other. 

Thus,  the  time  required  to  do  a  certain  amount  of  work  varies 
inversely  as  the  number  of  workmen  employed  ;  for,  if  the  number 
of  workmen  be  doubled,  halved,  or  changed  in  any  ratio,  the  time 
required  will  be  halved,  doubled,  or  changed  in  the  inverse  ratio. 

362.  If  A  varies  inversely  as  B,  two  values  of  A  have  to 
each  other  the  inverse  ratio  of  the  two  corresponding  values 
of  ^  ;  OT  a  :  a' : :  b' :  b,  that  is,  ab  =  a'b'. 


294  ALGEBRA. 

Hence,  the  product  AB  is  constant,  and  may  be  denoted 
by  m.     That  is,  AB  =  m. 

If  any  two  corresponding  values  of  A  and  B  are  known, 
the  constant  m  may  be  found. 

The  equation  AB^=m  may  be  written  A  =  —7  and  as  m 

is  constant,  A  is  said  to  vary  as  the  7'eciprocal  of  B,  or 

-I- 

363.  The  two  equations, 

A  =  mB  (for  direct  variation), 
^=:—  (for  inverse  variation), 

furnish  the  simplest  method  of  treating  Variation. 
If  A=^mBC,  A  is  said  to  ys^vj  jointly  as  B  and  C. 

If  ^  =  —7-'  A  is  said  to  vary  directly  as  B  and  inversely 
as  C. 

364.  The  following  results  are  to  be  observed  : 

I.  liA^B  and  ^cc  (7,  then  ^oc  (7. 

For  ^  =  m_B,  where  m  is  constant, 

and  B=nC,   where  71  is  constant. 

.-.A  =  mnC. 
.-.  A  cc  C,  since  mn  is  constant. 

In  like  manner,  if  Ace  B  and  Bcc—^  then  Ace—' 

Li  G 

II.  If  ^cc  Cand^QC  C,thenA±B^  (7,  and  V"Zeoc  a 

For  A  =  mC,  where  m  is  constant, 

and  B  =  nC,   where  n  is  constant. 

.'.A±B={m±n)  C. 

.:A±BacC,  since  m  ±  n  is  constant. 
Also,      Vl^  =  \/mCX  nC  =  VnmC'^  =  C 

.:  VAB  oc  C,  since  V  mn  is  constant. 


VARIATION.  295 

III.  If  J  Qc  i?  and  C  Qc  D,  then  AC  cc  BD. 

For  A  =  inB^  where  m  is  constant, 

C  =  nD^  where  n  is  constant. 
.\AC=  mnLD. 
.:AC=<x:  BD,  since  mn  is  constant. 

IV.  liA^B  then  A" cc  B". 

For  A  =  mB,  where  m  is  constant. 

.•.A"<x  B",  since  m'»  is  constant. 

V.    li  A^  B  when  C  is  unchanged,  and  A^  C  when  J5 
is  unchanged,  then  A°c  BC  when  both  B  and  6'  change. 

For        A  =  7?i5,  when  i?  varies  and  C  is  constant. 
Here,  m  is  constant  and  cannot  contain  the  variable  B. 
.:  A  must  contain  B,  but  no  other  power  of  B. 
Again,  A  =  nC,  when  C  varies  and  B  is  constant. 
Here,  n  is  constant  and  cannot  contain  the  variable  C, 
.'.A  must  contain  C,  but  no  other  power  of  C. 
Hence,  A  contains  both  B  and  C,  but  no  other  powers  of  B  and 
C,  and  therefore, 

A 

-^7-  =  p,  or  ^  =  pJ5C,  where  p  is  constant. 
.-.  ^  Qc  JBC,    since  p  is  constant. 


In  like  manner,  it  may  be  shown  that  if  A  varies  as  each 
of  any  number  of  quantities  B,  C,  D,  etc.,  when  the  rest 
are  unchanged,  then  when  they  all  change,  A  qc  BCD,  etc. 

Thus,  the  area  of  a  rectangle  varies  as  the  base  when  the  altitude 
is  constant,  and  as  the  altitude  when  the  base  is  constant,  but  as  the 
product  of  the  base  and  altitude  when  both  vary. 

The  volume  of  a  rectangular  solid  varies  as  the  length  when  the 
width  and  thickness  remain  constant ;  as  the  width  when  the  length 
and  thickness  remain  constant ;  as  the  thickness  when  the  length  and 
width  remain  constant ;  but  as  the  product  of  length,  breadth,  and 
thickness  when  all  three  vary. 


296  ALGEBRA. 

(1)  If  A  varies  inversely  as  B,  and  when  A  =  2  the  corre- 

sponding value  of  B  is  36,  find  the  corresponding 
value  of  B  when  A  =  9. 

Here  A  =  —^ 

n 

or  m  =  AB. 

.-.  m  =  2  X  36  =  72. 
And  if  9  and  72  are  substituted  for  A  and  m  respectively  in 

72 
the  result  is    9  =  —  • 

x> 
.-.95=  72. 
.-.  ^  =  8.   Arts. 

(2)  The  weight  of  a  sphere  of  given  material  varies  as  its 

volume,  and  its  volume  varies  as  the  cube  of  its 
diameter.  If  a  sphere  4  inches  in  diameter  weighs 
20  pounds,  find  the  weight  of  a  sphere  5  inches 
in  diameter. 

Let     W  represent  the  weight, 

V  represent  the  volume, 

D  represent  the  diameter. 
Then  TTx  V  and  Foe  L^. 

Put     W  =  mD^, 
then,  since  20  and  4  are  corresponding  volumes  of  W  and  D, 
20  =  m  X  64, 

.-.  when  D  =  b,W=j%  of  125  =  39xV. 

Exercise  115. 

1.  If  ^oc  B,  and  ^  =  4  when  B  =  5,  find  A  when  i?=:12. 

2.  If  ^  QC  B,  and  when  B  =  i,  A  =  i,  find  A  when  J5  =  -J-. 

3.  If  A  varies  jointly  as  B  and  C,  and  3,  4,  5  are  simultane- 

ous values  of  A,  B,  C,  find  A  when  B=  C  =  10, 


VARIATION.  297 

4.  li  A  cc  —,  and  B  =  2,  when  ^  =  10,  find  the  value  of 

X) 

B  when  A  =  4:. 

5.  If  A  oc  — ,  and  B  =  4:,  and  C=S,  when  ^  =  6,  find  the 

G 

value  of  A  when  B  =  5  and  C=7. 

6.  If  the  square  of  X  varies  as  the  cube  of  Y,  and  X=  3 

when  F=4,  find  the  equation  between  X  and  Y. 

7.  If  the  square  of  X  varies  inversely  as  the  cube  of  Y, 

and  X=  2  when  Z  =  3,  find  the  equation  between 
Xand  Y. 

8.  If  Z  varies  as  X  directly  and  F  inversely,  and  if  X=  3, 

and  y=4,  when  Z=2,  find  the  value  of  Z  when 
X=15and  Y=H. 

9.  If  Ace  B-\-c  where  c  is  constant,  and  if  ^  =  2  when 

B  =  l,  and  if  A=5  when  B  =  2,  find  ^  when  i?  =  3. 

10.  The  velocity  acquired   by  a  stone  falling  from  rest 

varies  as  the  time  of  falling ;  and  the  distance  fallen 
varies  as  the  square  'of  the  time.  If  it  is  found 
that  in  3  seconds  a  stone'  has  fallen  145  feet,  and 
acquired  a  velocity  of  96|-  feet  per  second,  find  the 
velocity  and  distance  at  the  end  of  5  seconds. 

11.  If  a  heavier  weight  draw  up  a  lighter  one  by  means 

of  a  string  passing  over  a  fixed  wheel,  the  space 
described  in  a  given  time  will  vary  directly  as  the 
difference  between  the  weights,  and  inversely  as 
their  sum.  If  9  ounces  draw  7  ounces  through  8 
feet  in  2  seconds,  how  high  will  12  ounces  draw  9 
ounces  in  the  same  time  ? 

12.  The  space  will  vary  also  as  the  square  of  the  time. 

Find  the  space  in  Example  11,  if  the  time  in  the 
latter  case  is  3  seconds. 


298  ALGEBRA. 

13.  Equal  volumes  of  iron  and  copper  are  found  to  weigh 

77  and  89  ounces  respectively.  Find  the  weight  of 
10^  feet  of  round  copper  rod  when  9  inches  of  iron 
rod  of  the  same  diameter  weigh  31y^o  ounces. 

14.  The  square  of  the  time  of  a  planet's  revolution  varies 

as  the  cube  of  its  distance  from  the  sun.  The 
distances  of  the  Earth  and  Mercury  from  the  sun 
being  91  and  35  millions  of  miles,  find  in  days  the 
time  of  Mercury's  revolution. 

15.  A  spherical  iron  shell  1  foot  in  diameter  weighs  ^\i^ 

of  what  it  would  weigh  if  solid.  Find  the  thickness 
of  the  metal,  it  being  known  that  the  volume  of  a 
sphere  varies  as  the  cube  of  its  diameter. 

16.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diam- 

eter. Compare  the  volume  of  a  sphere  6  inches  in 
diameter  with  the  sum  of  the  volumes  of  3  spheres 
whose  diameters  are  3,  4,  5  inches  respectively. 

17.  Two  circular  gold  plates,  each  an  inch  thick,  the  diam- 

eters of  which  are  6  inches  and  8  inches  respectively, 
are  melted  and  formed  into  a  single  circular  plate 
1  inch  thick.  Find  its  diameter,  having  given  that 
the  area  of  a  circle  varies  as  the  square  of  its 
diameter. 

18.  The  volume  of  a  pyramid  varies  jointly  as  the  area  of 

its  base  and  its  altitude.  A  pyramid,  the  base  of 
which  is  9  feet  square,  and  the  height  of  which  is 
10  feet,  is  found  to  contain  10  cubic  yards.  What 
must  be  the  height  of  a  pyramid  upon  a  base  3  feet 
square,  in  order  that  it  may  contain  2  cubic  yards  ? 


CHAPTER   XXI. 
Series. 

366.  A  SUCCESSION"  of  numbers  which  proceed  according 
to  some  fixed  law  is  called  a  series;  and  the  successive 
numbers  are  called  the  terms  of  the  series. 

Thus,  by  executing  the  indicated  division  of ,  the  series  1  +x 

X        X 

+  x^  +  x^  ■{■ is  obtained,  a  series  that  has  an  unlimited  number 

of  terms. 

366.  A  series  that  is  continued  indefinitely  is  called  an 
infinite  series ;  and  a  series  that  comes  to  an  end  at  some 
particular  term  is  called  a  finite  series. 

367.  When  x  is  less  than  1,  the  more  terms  we  take  of 

the  infinite  series  l-{-x-\-x^-\-x^-\- ,  obtained  by  dividing 

1  by  1— ic,  the  more  nearly  does  their  sum  approach  to  the 

value  of 

l  —  x 

1  13 

Thus,  if  X  =  ^,  then = =  -  ,  and  the  series  becomes 

1      X      1      -J-      2 

l  +  i+  1+  ^j-\- ,  a  sum  which  cannot  become  equal  to  |  how- 
ever great  the  number  of  terms  taken,  but  which  may  be  made  to 
differ  from  |  by  as  little  as  we  please  by  increasing  indefinitely  the 
number  of  terms. 

368.  But  when  x  is  greater  than  1,  the  more  terms  we 
take  of  the  series  1 -\- x -\- x"^ '\- x^ -{- the  more  does  the 

sum  of  the  series  diverge  from  the  value  of     ^    * 

JL  *""  X 


1       _        1 

-3""      2 
1  +  3  +  9  +  27  + ,a  sum  which  diverges  more  and  more  from  —  |, 


Thus,  if  X  =  3,  then  ; = =  —  - ,  and  the  series  becomes 

'  1  —  X      1  —  3  2 


300  ALGEBRA. 

the  more  terms  we  take,  and  which  may  be  made  to  increase  indefi- 
nitely by  increasing  indefinitely  the  number  of  terms  taken. 

369.  A  series  whose  sum  as  the  number  of  its  terms  is 
indefinitely  increased  approaches  some  fixed  finite  value  as 
a  limit  is  called  a  converging  series;  and  a  series  whose 
sum  increases  indefinitely  as  the  number  of  its  terms  is 
increased,  is  called  a  diverging  series. 

370.  When  x  =  l,  the  division  of  Ihj  1—x,  that  is,  of 
1  by  0,  has  no  meaning,  according  to  the  definition  of  divis- 
ion; and  any  attempt  to  divide  by  a  divisor  that  is  equal 
to  zero  leads  to  absurd  results. 

Thus,  8  +  4=8  +  4; 

by  transposing,  8  —  8  =  4  —  4; 

or,  dividing  by  4  —  4,         2=1;         a  manifest  absurdity. 

371.  When  x=^l  very  nearly,  then  the  value  of  ;;; 

1  —  x 

will  be  very  great,  and  the  sum  of  the  series  1 -{- x -\- x'^ -{- 

x^~\- will  become  greater  and  greater  the  more  terms  we 

take.     Hence,  by  making  the  denominator  1  —  x.  approach 

indefinitely  to  zero,  the  value  of  the  fraction  ~ may  be 

made  to  increase  at  pleasure. 

372.  If  the  symbol  o  be  used  to  denote  a  number  that 
is  less  than  any  assignable  number,  and  that  may  be  con- 
sidered to  decrease  without  limit,  not,  however,  becoming 
0,  and  the  symbol  go  be  used  to  denote  a  number  that  is 
greater  than  any  assignable  number,  and  that  may  be 
considered  to  increase  without  limit,  not,  however,  becom- 
ing 00,  then  ^ 

=  00. 

o       — 
In  the  same  sense  — -=  g^,  where  a  represents  any  value 
that  may  be  assigned. 


SERIES.  301 


373.  If  X  in  the  fraction  -^j- —  be  equal  to  1,  the  numer- 
ator  and  denominator  will  each  become  0,  and  the  fraction 
will  assume  the  form  —  • 

374.  If,  however,  x  in  this  fraction  approach  to  1  as  its 
limit,  then  the  denominator  1  —  x,  inasmuch  as  it  has  some 
value,  even  though  less  than  any  assignable  value,  may  be 
used  as  a  divisor,  and  the  result  is  1 -\- x -\- x^ -\- x^ -{- x^. 
Hence,  it  is  evident  that  though  both  terms  of  the  fraction 
become  smaller  and  smaller  as  1  —  x  approaches  to  0,  still 
the  numerator  becomes  more  and  more  nearly  Jive  times 
the  denominator. 

It  may  be  remarked  that  when  the  symbol  §  is  obtained  for  the 
value  of  the  unknown  number  in  a  problem,  the  meaning  is  that  the 
problem  has  no  d^nite  solution,  but  that  its  conditions  are  satisfied 
if  any  value  whatever  be  taken  for  the  required  number ;  and  if  the 
symbol  5,  in  which  a  denotes  any  assigned  value,  is  obtained  for  the 
value  of  the  unknown  number,  the  meaning  is  that  the  conditions  of 
the  problem  are  impossible. 

375.  The  number  of  different  series  is  unlimited,  but  the 
only  kind  of  series  that  will  be  considered  at  this  stage  of 
the  work  are  Arithmetical,  Geometrical,  and  Harmonical 
Series. 

Arithmetical  Series. 

376.  A  series  in  which  the  difference  between  any  two 
adjacent  terms  is  equal  to  the  difference  between  any  other 
two  adjacent  terms,  is  called  an  Arithmetical  Series  or  an 
Arithmetical  Progression. 

377.  The  general  representative  of  such  a  series  will  be 

a,  a-^d,  a-\-2d,  a-\-^d , 

in  which  a  is  the  first  term  and  d  the  common  difference  j 


302  ALGEBRA. 

and  the  series  will  be  increasing  or  decreasing  according  as 
d  is  positive  or  negative. 

378.  Since  each  succeeding  term  of  the  series  is  obtained 
by  adding  d  to  the  preceding  term,  the  coefficient  of  d  will 
always  be  1  less  than  the  number  of  the  term,  so  that  the 

nth  term  =  a-\-(n  —  1)  d. 

If  the  nth  term  is  denoted  by  I,  this  equation  becomes 

l  =  a-\-{n  —  l)d.  (1) 

379.  The  arithmetical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
an  arithmetical  series. 

380.  If  a  and  h  denote  two  numbers,  and  A  their  arith- 
metical mean,  then,  by  the  definition  of  an  arithmetical 
series, 

A  —  a  =  h  —  A. 

•••^  =  ^-  (2) 

381.  Sometimes  it  is  required  to  insert  several  arith- 
metical means  between  two  numbers. 

If  m  =  the  number  of  means,  then  m-\-2=^n,  the  whole 

number  of  terms ;  and  ii  m-\-2  is  substituted  for  n  in  the 

equation 

l  =  a-\-(n  —  1)  tZ, 

the  result  is  l=a-\-(m-\-V)d, 

By  transposing  a,       I  —  a  =  {m  -f- 1)  ^. 

I  —  a 
.-. r^  =  d.  (3) 

Thus,  if  it  is  required  to  insert  six  means  between  3  and  17,  the 

Yi 3 

value  of  d  is  found  to  be  =  2  ;  and  the  series  will  be  3,  5,  7,  9, 

11,  13,  15,  17. 


SERIES.  303 

382.  If  I  denotes  the  last  term,  a  the  first  term,  n  the 
number  of  terms,  d  the  common  difference,  and  s  the  sum 
of  the  terms,  it  is  evident  that 

s=      a     -\-(a-\-d)-{-(a-\-2d)-\- -{-(l—d)  +      I,     or 

s=:      I     J^(l—d)J^ll  —  2d.)-\- +  (a+^)4-      a 

,'.2s  =  (a-^l)^{a^l)   +(a+0      + +  (a+Z)+(a+0 

=  n(a-\-l^ 

.•.«  =  ^(«  +  0-  (4) 

383.  From  the  two  equations, 

l=,a-\-{n  —  l)d,  (1) 

«  =  2^'^  +  ^'  (2) 

any  one  of  the  numbers  a,  d,  I,  n,  s  may  be  found  when 
three  are  given. 

Ex.   Find  n  when  d,  I,  s  are  given. 

From  (1),  a=l—{n—l)d. 

T^       /ox                                2  s  — In 
From  (2),  a  = 

Therefore,   I- (n- l)d=^^~~^'^' 

.-.  In  —  dn'^-\-dn  =  2s  —  In. 
.'.  d?i2  -  {2l+d)n=—2s. 
.•.4d%2_()  +  (2Z  +  d)2=  (2Z  +  (i)2-8(Zs, 

.-. 2dn-{2l+d)  =  ±  V(2 Z  +  d)^  - 8 ds. 

_2l  +  d  ±\/{2l  -\-  d)-^  -  Sds 
•'•''~  2d 

Note.  The  table  on  the  following  page  contains  the  results  of  the 
general  solution  of  all  possible  problems  in  arithmetical  series.  The 
student  is  advised  to  work  these  out,  both  for  the  results  obtained 
and  for  the  practice  gained  in  solving  literal  equations  in  which  the 
imknown  numbers  are  represented  by  other  letters  than  x,  y,  z. 


304 


ALGEBRA. 


No. 

Given. 

Required. 

Results. 

1 
2 
3 
4 

a  d  n 
ads 
a  n  s 
d  n  s 

I 

l=a+  {n-l)d. 

l=-id±\/2ds+{a-id)^. 

1  = a. 

n 

^_s       (n-l)d. 

n            2 

5 
6 

7 
8 

a  d  n 
a  d  I 
a  n  I 
d  n  I 

s 

s  =  in[2a+  {n  —  l)d]. 

l  +  a^l^-a^ 
'=     2     ^     2d    ' 

s  =  in[2l-{n-l)d]. 

9 
10 
11 
12 

d  n  I 
d  n  s 
d  I  s 
n  I  s 

a 

a=l-  {n-\)d. 
s       (n  -l)d 

a  =  i(Z±V(Z  +  id)2-2ds. 

2s      , 

a  = I. 

n 

13 
14 
15 
16 

a  n  I 
a  n  s 
a  I  s 
n  I  s 

d 

n  —  1 
2(s  — an) 
~  n  (n  -  1) 
Z2  -  a2 
^  =  2s-l~a 

2(nZ-s). 
'^~  n{n-\) 

17 
18 
19 
20 

a  d  I 
ads 
a  I  s 
d  I  s 

n 

^=     cZ     +^• 

cZ  -  2  a  ±  V(2  a  -  d)2  +  8  ds 

2d 

2s 
n  =  ,  , 
l  +  a 

2Z  +  d±V(2Z+d)2-8ds 

SERIES.  305 


Exercise  116. 


Find  the  thirteenth  term  of  5,  9,  13 

ninth  term  of  — 3,  — 1,  1 

tenth  term  of  —2,  —5,  —8 

eighth  term  of  a,  a-\-3b,  a -{-6b 

fifteenth  term  of  1,  f ,  f 

thirteenth  term  of  —  48,  —  44,  —  40. 


2.  The  first  term  of  an  arithmetical  series  is  3,  the  thir- 

teenth term  is  55.     Find  the  common  difference. 

3.  Find  the  arithmetical  mean  between  :  («.)  3  and  12  ; 

(b.)  —  5  and  17  ;   (c.)  a^  +  ab  —  b^  and  a^  —  ab  +  b\ 

4.  Insert  three  arithmetical  means  between  1  and  19 ; 

and  four  arithmetical  means  between  — 4  and  17. 

5.  The  first  term  of  a  series  is  2,  and  the  common  differ- 

ence ^.     What  term  will  be  10  ? 

6.  The  seventh  term  of  a  series,  whose  common  difference 

is  3,  is  11.     Find  the  first  term. 

7.  Find  the  sum  of 

5  +  8  + 11  -f- to  ten  terms. 

—  4  —  1  +  2  + to  seven  terms. 

a-\-4ia-\-l<i-\r to  n  terms. 

§  +  -j^y  -|-  ^\  + to  twenty-one  terms. 

1  -f  2|-  +  4^-  + to  twenty  terms. 

8.  The  sum  of  six  numbers  of  an  arithmetical  series  is 

27,  and  the  first  term  is  1.     Determine  the  series. 

9.  How  many  terms  of  the  series  —  5  —  2  +  1  + must 

be  taken  so  that  their  sum  may  be  63  ? 

10.    The  first  term  is  12,  and  the  sum  of  ten  terms  is  10. 
Find  the  last  term. 


306  ALGEBRA. 

11.  The  arithmetical  mean  between  two  numbers  is  10, 

and  the  mean  between  the  double  of  the  first  and 
the  triple  of  the  second  is  27.     Eind  the  numbers. 

12.  Find  the  middle  term  of  eleven  terms  whose  sum  is  66. 

13.  The  first  term  of  an  arithmetical  series  is  2,  the  com- 

mon difference  is  7,  and  the  last  term  79.  Find  the 
number  of  terms. 

14.  The  sum  of  15  terms  of  an  arithmetical  series  is  600, 

and  the  common  difference  is  5.     Find  the  first  term. 

15.  Insert  ten  arithmetical  means  between  — 7  and  114. 

16.  The  sum  of  three  numbers  in  arithmetical  progression 

is  15,  and  the  sum  of  their  squares  is  83.     Find  the 

numbers. 

Let  X  —  y,  X,  X  +  y  represent  the  numbers. 

17.  Arithmetical  means  are  inserted  between  5  and  23,  so 

that  the  sum  of  the  first  two  is  to  the  sum  of  the  last 
two  as  2  is  to  5.      How  many  means  are  inserted  ? 

18.  Find  three  numbers  of  an  arithmetical  series  whose 

sum  shall  be  21,  and  the  sum  of  the  first  and  second 
shall  be  f  of  the  sum  of  the  second  and  third. 

19.  Find  three  numbers  whose  common  difference  is  1, 

such  that  the  product  of  the  second  and  third  ex- 
ceeds the  product  of  the  first  and  second  by  -J-. 

20.  How  many  terms  of  the  series  1,  4,  7 must  be  taken, 

in  order  that  the  sum  of  the  first  half  may  bear  to 
the  sum  of  the  second  half  the  ratio  10  :  31  ? 

21.  A  travels  uniformly  20  miles  a  day ;  B  starts  three 

days  later,  and  travels  8  miles  the  first  day,  12  the 
second,  and  so  on,  in  arithmetical  progression.  In 
how  many  days  will  B  overtake  A  ? 


SERIES.  307 

22.  A  number  consists  of  three  digits  which  are  in  arith- 

metical progression  j  and  this  number  divided  by 
the  sum  of  its  digits  is  equal  to  26  ;  but  if  198  be 
added  to  it,  the  digits  in  the  units'  and  hundreds' 
places  will  be  interchanged.     Find  the  number. 

23.  The  sum  of  the  squares  of  the  extremes  of  four  num- 

bers in  arithmetical  progression  is  200,  and  the  sum 
of  the  squares  of  the  means  is  136.  What  are  the 
numbers  ? 

24.  Show  that  if  any  even  number  of  terms  of  the  series 

1,  3,  5 is  taken,  the  sum  of  the  lirst  half  is  to 

the  sum  of  the  second  half  in  the  ratio  1  : 3. 

25.  A  and  B  set  out  at  the  same  time  to  meet  each  other 

from  two  places  343  miles  apart.  Their  daily  jour- 
neys were  in  arithmetical  progression,  A's  increase 
being  2  miles  each  day,  and  B's  decrease  being  5 
miles  each  day.  On  the  day  at  the  end  of  which 
they  met,  each  travelled  exactly  20  miles.  Find 
the  duration  of  the  journey. 

26.  Suppose  that  a  body  falls  through  a  space  of  16^  feet 

in  the  first  second  of  its  fall,  and  in  each  succeeding 
second  32^  more  than  in  the  next  preceding  one. 
How  far  will  a  body  fall  in  20  seconds  ? 

27.  The  sum  of  five  numbers  in  arithmetical  progression 

is  45,  and  the  product  of  the  first  and  fifth  is  |  of 
the  product  of  the  second  and  fourth.  Find  the 
numbers. 

28.  If  a  full  car  descending  an  incline  draw  up  an  empty 

one  at  the  rate  of  1-J-  feet  the  first  second,  4-J-  feet 
the  next  second,  7^  feet  the  third,  and  so  on,  how 
long  will  it  take  to  descend  an  incline  150  feet  in 
length  ?  What  part  of  the  distance  will  the  car 
have  descended  in  the  first  half  of  the  time  ? 


308  ALGEBBA. 


Geometrical  Series. 

384.  A  series  is  called  a  Geometrical  Series  or  a  Geomet- 
rical Progression  when  each  succeeding  term  is  obtained  by 
multiplying  the  preceding  term  by  a  constant  multiplier. 

385.  The  general  representative  of  such  a  series  will  be 

a,  ar,  at^,  ai^,  ar* , 

in  which  a  is  the  first  term  and  r  the  constant  multiplier 
or  ratio. 

386.  Since  the  exponent  of  r  increases  by  1  for  every 
term,  the  exponent  will  always  be  1  less  than  the  number 
of  the  term ;  so  that  the 

nth  term  =  ar"~^. 

387.  If  the  nth  term  is   denoted  by  I,  this    equation 

becomes 

l=ai^-\  (1) 

388.  The  geometrical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
a  geometrical  series. 

389.  If  a  and  b  denote  two  numbers,  and  G  their  geo- 
metrical mean,  then,  by  definition  of  a  geometrical  series, 


G_ 

a 

b 

.'.  G  = 

-.ylah 

(2) 

390.    Sometimes  it  is  required  to  insert  several  geomet- 
rical means  between  two  numbers. 


SERIES.  309 

li  m  =  the  number  of  means,  then  m  -f  2  =  n,  the  whole 
number  of  terms  ;  and  if  m  +  2  is  substituted  for  n  in  the 
equation  i  =  ar"~^ 

the  result  is  1  =  ar'^'^K 

.-.7^+1  =  -  (3) 

Thus,  if  it  is  required  to  insert  three  geometrical  means  between 
3  and  48,  the  value  of  r  is  found  to  be 

and  the  series  will  be         3,  6,  12,  24,  48. 

391.  If  I  denotes  the  last  term,  a  the  first  term,  n  the 
number  of  terms,  r  the  common  ratio,  and  s  the  sum  of  the 
n  terms,  then 

5  =  a  -f  ar  +  o.^^  +  «^  + +  «?'"~^ 

Multiply  by  r,  rs  =  ar  -\-  ai^  -\-  ai^  + -|-  a/*"~^  -f-  a?*". 

Therefore,  by  subtracting  the  first  equation  from  the 
second,  rs  —  s  =  ar"  —  a, 

or  (r—l)s  =  a  (/•"  —  !). 

a  (;•«  —  1) 

392.  When  r  is  less  than  1,  this  formula  will  be  more 
convenient  if  written 

1  —  r 

393.  Since  l=a7^-\ 
then                                        rl=ar^, 

and  (4)  may  be  written       s=     _    • 

In  working  out  the  following  results,  the  student  will  make  use  of 

a  if"  —  1) 
the  two  equations,  I  =  ar^-^  and  s  =     r-~\ 


310 


ALGEBRA. 


No. 
1 
2 
3 
4 


Given. 


Required. 


Results. 


a  r  n 
a  r  s 
a  n  s 
r  n  s 


a  r  n 
a  r  I 
a  n  I 
r  n  I 


1  = 
1  = 
l{s 
1  = 


a+  {r-l)s 

r 
—  0"-'  —  a{s  —  a)»-i  =  0. 
(r  —  1)  sr»-^ 
r«—  1 


a(r''  — 1) 

r-  1 
rl  —  a 
r-  l' 


n— 1 


a» 


»j— 1 ,7-       n— 1  /— ■ 


^r»  —  ^ 
y  ?i  —  yi  ?i — 1 


10 
11 
12 


r  n  Z 

r  n  s 

r  I  s 

n  I  s 


I 

fU—l 

ir-l)s_ 

W  —  (r  —  1)  s. 

-  a)"-i  -  Z  (s  -  Z)»-i  =  0. 


13 
14 
15 
16 

17 
18 
19 
20 


a  n  I 

a  n  s 

a  I  s 

n  I  s 


_  «-i  // 


j-n  . 


s      .  s—a      - 

-  -  r  H =  0. 

a  a 

s  —  a 

s-  i 

--^,r"-i+-^  =  0. 
s—  I  s  —  I 


a  r  I 
a  r  s 
a  I  8 
r  I  s 


log  I  —  loff  a 


+  1. 


log  r 

log  [a  +  (r  —  1)  g]  —  log  a 
logr 
log  ?  —  log  g  J   ^ 

log(s  — a)  — log(s  — Z) 
logZ-log[Zr—  (r— l)s]   ^   ^ 
I02;  r 


SERIES.  311 


Exercise  117. 


1.    Find  the  seventh  term  of  2,  6,  18 

sixth  term  of  3,  6,  12 

ninth  term  of  6,  3,  1^ 

eighth  term  of  1,  ~2,  4 

twelfth  term  of  x^,  x'^,  x^ 

fifth  term  of  4  a,  —  6  ma^,  9  7/i^a^. 


2.  Find  the  geometrical  mean  between  ISx^y  and  30  xi/^z. 

3.  Find  the  ratio  when  the  first  and  third  terms  are  5 

and  80  respectively. 

4.  Insert  two  geometrical  means  between  8  and  125 ;  and 

three  between  14  and  224. 

5.  If  ar=2  and  r  =  3,  which  term  will  be  equal  to  162? 

6.  The  fifth  term  of  a  geometrical  series  is  48,  and  the 

ratio  2.     Find  the  first  and  seventh  terms. 

7. 


Find  the  sum  of 

3  +  6  +  12+ 

to  eight  terms. 

1-3  +  9- 

to  seven  terms. 

8  +  4  +  2+ 

to  ten  terms. 

0.1 +0.5 +2.5+- 

••  to  seven  terms. 

to  five  terms. 

8.  The  population  of  a  city  increases  in  four  years  from 

10,000  to  14,641.     What  is  the  rate  of  increase  ? 

9.  The  sum  of  four  numbers  in  geometrical  progression 

is  200,  and  the  first  term  is  5.     Find  the  ratio. 

10,    Find  the  sum  of  eight  terms  of  a  series  whose  last 
term  is  1,  and  fifth  term  ^. 


312  ALGEBRA. 

11.  In  an  odd  number  of  terms,  show  that  the  product  of 

the  first  term  and  last  term  will  be  equal  to  the 
square  of  the  middle  term. 

12.  The  product  of  four  terms  of  a  geometrical  series  is  4, 

and  the  fourth  term  is  4.     Determine  the  series. 

13.  If  from  a  line  one-third  be  cut  off,  then  one-third  of 

the  remainder,  and  so  on,  what  fraction  of  the  whole 
will  remain  when  this  has  been  done  five  times  ? 

14.  Of  three  numbers  in  geometrical  progression,  the  sum 

of  the  first  and  second  exceeds  the  third  by  3,  and 
the  sum  of  the  first  and  third  exceeds  the  second 
by  21.     What  are  the  numbers  ? 

15.  Find  two  numbers  whose  sum  is  3^  and  geometrical 

mean  l-J. 

16.  A  glass  of  wine  is  taken  from  a  decanter  that  holds  ten 

glasses,  and  a  glass  of  water  poured  in.  After  this 
is  done  five  times,  what  part  of  the  contents  is  wine  ? 

17.  There  are  four  numbers  such  that  the  sum  of  the  first 

and  the  last  is  11,  and  the  sum  of  the  others  is  10. 
The  first  three  of  these  four  numbers  are  in  arith- 
metical progression,  and  the  last  three  are  in  geo- 
metrical progression.     Find  the  numbers. 

18.  Find  three  numbers  in  geometrical  progression  such 

that  their  sum  is  13  and  the  sum  of  their  squares 
is  91. 

19.  The  difference  between  two  numbers  is  48,  and  the 

arithmetical  mean  exceeds  the  geometrical  mean  by 
18.     Find  the  numbers. 

20.  There  are  four  numbers  in  geometrical  progression, 

the  second  of  which  is  less  than  the  fourth  by  24, 
and  the  sum  of  the  extremes  is  to  the  sum  of  the 
means  as  7  to  3.     Find  the  numbers. 


SERIES.  313 

21.  A  number  consists  of  three  digits  in  geometrical  pro- 

gression. The  sum  of  the  digits  is  13 ;  and  if  792 
be  added  to  the  number,  the  digits  in  the  units'  and 
hundreds'  places  will  be  interchanged.  Mnd  the 
number. 

394.  When  r  is  less  than  1,  a  geometrical  series  has  its 
terms  continually  decreasing;  and  by  increasing  n,  the 
value  of  the  nth  term,  at^-^,  may  be  made  as  small  as  we 
please,  though  not  absolutely  zero. 

395.  The  formula  for  the  sum  of  n  terms, 

a(l— r") 
1-r 

may  be  written  :; -. 

•^  \  —  r      1  —r 

By  increasing  n  indefinitely,  the  value  of     becomes 

indefinitely  small,  so  that  the  sum  of  n  terms  approaches 
indefinitely  to     _     as  its  limit. 

Ex.    rind  the  limit  of  1  —  ^  +  i  —  -J  + 

Here  a  =  1,  and  r  =  —  i, 

and  therefore  the  limit  :; =  :; ; — r-  =  rTTT  —  o'  ^^* 

1  —  r      1  —  (—  i)       1  +  i      3 

22.  Find  the  limits  of  the  sums  of  the  following  infinite 

series : 

4  +  2  +  1+ 2-li  +  f- 

^  +  i  +  f  + 0.1  +  0.01  +  0.001  + 

i-TV  +  ^4- 0.868686 

1-1  +  ^^- 0.54444 

1  +  tV  +  .V+ 0.83636 


314  ALGEBRA. 

*  Harmonical  Series. 

396.  A  series  is  called  a  Harinonical  Series,  or  a  Har- 
monica! Progression,  when  the  reciprocals  of  its  terms  form 
an  arithmetical  series. 

Hence,  the  general  representative' of  such  a  series  will  be 

1^       1  1  ^  1 

a     a-\-d     a-\-2d  a-^(n — l)d 

397.  Questions  relating  to  harmonical  series  should  be 
solved  by  writing  the  reciprocals  of  its  terms  so  as  to  form 
an  arithmetical  series. 

398.  If  a  and  b  denote  two  numbers,  and  H  their  har- 
monical mean,  then,  by  the  definition  of  a  harmonical  series, 

R      a~  b      H 

.  2  _1      1_^  +  ^. 
'''  H~  a^  b~    ab    ' 

2ab 


r.II 


a-^b 


399.    Sometimes  it  is  required  to  insert  several  harmon- 
ical means  between  two  numbers. 

Ex.    Let  it  be  required  to  insert  three  harmonical  means 
between  3  and  18. 

Find  the  three  arithmetical  means  between  ^  and  Jg. 

These  are  found  to  be  }|,  f|,  /g ;   therefore,  the  harmonical  means 

QT-p   7  2      7  2      72.     r>r.^l5      Kl      Q 

*  A  harmonical  series  is  so  called  because  musical  strings  of  uniform 
thickness  and  tension  produce  harmony  when  their  lengths  are  repre- 
sented by  the  reciprocals  of  the  natural  series  of  numbers  ;  that  is,  by 
I    1    1 


SERIES.  315 

Exercise  118. 

1.  Insert  four  harmonical  means  between  2  and  12. 

2.  Find  two  numbers  whose  difference  is  8  and  the  har- 

monical mean  between  them  1|. 

3.  Find  the  seventh  term  of  the  harmonical  series  3,  3f , 

4..... 

4.  Continue  to  two  terms  each  way  the  harmonical  series 

two  consecutive  terms  of  which  are  15,  16. 

5.  The  first  two  terms  of  a  harmonical  series  are  5  and  6. 

Which  term  will  equal  30  ? 

6.  The  fifth  and  ninth  terms  of  a  harmonical  series  are 

8  and  12.     Find  the  first  four  terms. 

7.  The  difference  between  the  arithmetical  and  harmon- 

ical means  between  two  numbers  is  1|,  and  one  of 
the  numbers  is  four  times  the  other.  Find  the 
numbers. 

8.  Find   the   arithmetical,   geometrical,   and   harmonical 

means  between  two  numbers  a  and  h\  and  show 
that  the  geometrical  mean  is  a  mean  proportional 
between  the  arithmetical  and  harmonical  means. 
Also,  arrange  these  means  in  order  of  magnitude. 

9.  The  arithmetical  mean  between  two  numbers  exceeds 

the  geometrical  by  13,  and  the  geometrical  exceeds 
the  harmonical  by  12.     What  are  the  numbers  ? 

10.  The  sum  of   three  terms  of   a  harmonical  series  is 

11,  and  the  sum  of  their  squares  is  49.  Find  the 
numbers. 

11.  When  a,  i,  c  are  in  harmonical  progression,  show  that 

a'.C'.'.a  —  h  '.h  —  c. 


CHAPTER   XXII. 

*  Choice.     Binomial  Theorem. 

400.  If  three  paths,  A,  B,  and  C,  lead  to  the  top  of  a 
mountain,  there  is  obviously  a  choice  of  three  different 
ways  of  ascending  the  mountain  ;  and  when  the  top  of  the 
mountain  is  reached,  there  is  again  a  choice  of  three  differ- 
ent ways  of  descending. 

How  many  different  ways  are  there  of  doing  both  ? 

If  a  traveller  ascend  by  A,  he  may  descend  by  A,  B,  or 
C.  This  makes  three  ways  of  doing  both.  If  he  ascend 
by  B,  he  may  descend  by  A,  B,  ot  C -,  and  again,  if  he 
ascend  by  C,  he  may  descend  by  A,  B,  or  C. 

Therefore,  there  are  3X3  =  9  ways  in  all  of  doing  both. 

These  ways  may  be  indicated  as  follows  : 

1.  A  and  A.  4.  B  and  A.  7.    C  and  A. 

2.  A  and  B.  5.  B  and  B.  8.    C  and  B. 

3.  A  and  C.  6.  B  and  C.  9.   C  and  C. 

401.  Suppose  the  traveller  does  not  wish  to  ascend  and 
descend  by  the  same  path,  then  what  choice  has  he? 

He  has  a  choice  of  three  different  ways  in  ascending. 

But  he  has  a  choice  of  only  two  ways  in  descending. 

If  each  of  the  three  ways  of  ascending  be  joined  to  the 
two  eligible  ways  of  descending,  the  result  is  3X2  =  6 
ways  of  doing  both. 

*  This  chapter  is  based  upon  Whitworth's  Choice  and  Chance,  and 
many  of  the  examples  have  been  taken  from  that  elegant  work. 


CHOICE. 

These  ways  are : 

1.    A  and  B. 

3.    B  and  A. 

5.    C  and  ^. 

2.    A  and  C. 

4.    B  and  C. 

6.    6'and^. 

317 


402.  If  a  box  contains  five  capital  letters,  A,  B,  C,  Z>,  E, 
and  three  small  letters,  x,  y,  z,  in  how  many  different  ways 
may  a  capital  letter  and  a  small  letter  be  selected  ? 

A  capital  letter  may  be  selected  in  5  ways. 

With  each  capital  letter  selected,  a  small  letter  can  be 
joined  in  3  ways.  So  that  the  number  of  different  ways 
in  which  the  selection  can  be  made  is  3  X  5  =  15.     These 

ways  are : 

Ax         Bx         Cx         Dx         Ex 

Ay         By          Cy         By         Ey 
Az         Bz         Cz         Dz         Ez 

403.  Hence  the  fundamental  principle  of  choice : 

I.  If  one  thing  can  be  done  in  a  different  ways,  and  (when 
it  has  been  done  in  any  one  of  these  ways)  another  thing  can 
be  done  in  b  different  ways,  then  both  can  be  done  t/i  a  X  b 
different  ways. 

For,  corresponding  to  each  of  the  a  ways  of  doing  the  first  thing, 
there  are  h  ways  of  doing  the  second  thing.  Therefore,  altogether, 
there  are  a  X  6  ways  of  doing  both  things. 

(1)  On  a  shelf  are  7  English  and  5  French  books.     In  how 

many  ways  can  one  of  each  be  chosen? 
7  X  5  =  35.    Ans. 

(2)  On  a  shelf  are  7  English,  5  Trench,  and  9  German 

books.     In  how  many  ways  can  two  books  be  chosen 
so  that  they  shall  be  in  different  languages? 

An  English  book  and  a  French  book  can  be  chosen  in  7  X  5 
=  35  ways.  A  French  book  and  a  German  book  in  5  X  9  =  45 
ways.    An  English  book  and  a  German  book  in  7  X  9  =  63  ways. 

Hence,  there  is  a  choice  of  35  +  45  +  63  =  143  ways.    Atis. 


318  ALGEBRA. 

(3)  Out  of  8  different  pairs  of  gloves,  in  how  many  differ- 

ent ways  can  a  right-Land  and  a  left-hand  glove  be 
chosen  which  shall  not  form  a  pair  ? 

A  right-hand  glove  can  be  chosen  in  8  ways  ;  and  when  it 
is  chosen  there  are  7  left-hand  gloves,  any  one  of  which  may 
be  put  with  it  without  making  a  pair.  Hence,  the  choice  is  in 
8  X  7  =  50  ways. 

(4)  Out  of  the  ten  digits,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  how 

many  numbers  each  consisting  of  two  figures  can  be 
formed  ? 

Since  0  has  no  value  in  the  left-hand  place,  the  left-hand  place 
can  be  filled  in  9  ways. 

The  right-hand  place  can  be  filled  in  10  ways,  since  repetitions 
of  the  digits  are  allowed  (as  22,  33,  etc.). 

Hence,  the  whole  number  is  9  X  10  =  90. 

(5)  How  many  odd  numbers  consisting  of  two  figures  can 

be  formed  with  the  ten  digits  ? 

The  left-hand  place  can  be  filled  in  9  ways ;   the  right-hand 
place  in  only  5  ways,  since  it  must  be  either  1,  3,  5,  7,  or  9. 
Hence,  the  number  is,  9  X  5  =  45.    Ans. 

404.    By  a  simple  extension  of  Kule  I.  it  is  evident  that, 

II.  If  one  thing  can  he  done  in  a  ways,  and  then  a  second 
thing  can  he  done  in  b  ways,  then  a  third  in  c  ways,  then  a 
fourth  m  d  ways,  etc.,  the  numher  of  ways  of  doing  all  the 
things  will  ^e  a  X  b  X  c  X  d,  etc. 

(1)    In  how  many  ways  can   four  Christmas  presents  be 
given  to  four  boys,  one  to  each  boy  ? 

The  first  present  may  be  given  to  any  one  of  the  boys ;  hence 
there  are  4  ways  of  disposing  of  it. 

The  second  present  may  be  given  to  any  one  of  the  other  three 
boys ;  hence  there  are  3  ways  of  disposing  of  it. 

The  third  present  may  be  given  to  either  of  the  other  two 
boys ;  hence  there  are  2  ways  of  disposing  of  it. 


CHOICE.  319 

The  fourth  present  must  be  given  to  the  last  boy  ;  hence  there 
is  only  one  way  of  disposing  of  it. 

There  are,  then,  4x3x2x1  =  24  ways. 

(2)  In  how  many  ways  can  five  presents  be  given  to  two 

children  ? 

Each  present  may  be  disposed  of  in  2  ways ;  for  it  may  be 
given  to  either  child.  Hence,  the  whole  number  of  ways  of 
giving  the  presents  is2X2X2X2X2  =  32. 

(3)  In  how  many  ways  can  five  presents  be  divided  between 

two  chiklren  ? 

This  question  differs  from  the  last  only  in  the  fact  that  a 
division  of  the  gifts  excludes  the.  two  ways  in  which  either  child 
receives  all  the  gifts. 

Hence,  there  are  32  —  2  =  30  ways. 

(4)  In  how  many  ways  can  x  things  be  given  to  n  persons  ? 

n^.  Ans. 

(5)  In  how  many  ways  can  a  vowel  and  a  consonant  be 

chosen  out  of  the  alphabet  ? 

Since  there  are  in  the  alphabet  6  vowels  and  20  consonants, 
a  vowel  can  be  chosen  in  6  ways  and  a  consonant  in  20  ways, 
and  both  (Rule  I.)  in  6  X  20  =  120  ways. 

(6)  In  how  many  ways  can  a  two-lettered  word  be  made, 

containing  one  vowel  and  one  consonant  ? 

The  vowel  can  be  chosen  in  6  ways  and  the  consonant  in  20 
ways ;  and  then  each  combination  of  a  vowel  and  a  consonant 
can  be  written  in  2  ways  ;  as  ac,  ca. 

Hence,  the  whole  number  of  ways  is  0  X  20  X  2  =  240. 

405.  The  last  two  examples  show  the  difference  between 
a  selection  or  combination  of  different  things,  and  an 
arrangement  or  permutation  of  the  same  things. 


320  ALGEBRA. 

Thus,  ac  forms  a  combination  of  a  vowel  and  a  consonant,  and  ac 
and  ca  form  two  different  arrangements  of  this  combination. 

From  (5)  it  is  seen  that  120  different  combinations  can  be  made 
with  a  vowel  and  a  consonant ;  and  from  (6)  it  is  seen  that  240  differ- 
ent permutations  can  be  made  with  the  same. 

Again,  a,  6,  c  is  a  selection  of  three  letters  from  the  alphabet. 
This  selection  then  admits  of  6  different  arrangements  or  permuta- 
tions, as  follows : 

abc  bca  cab 

acb  bac  cba 

406.  A  selection  or  combination  of  any  number  of  ele- 
ments or  things  means  a  group  of  that  number  of  elements 
or  things  put  together  without  regard  to  their  order  of 
sequence.  An  arrangement  or  permutation  of  any  number 
of  elements  or  things  means  a  group  of  that  number  of 
elements  or  things  put  together  with  reference  to  their 
order  of  sequence. 

Arrangements  or  Permutations. 

407.  In  how  many  ways  can  the  letters  of  the  word 
Cambridge  be  arranged,  taken  all  at  a  time  ? 

There  are  nine  letters.  In  making  any  arrangement  any  one  of 
the  letters  may  be  put  in  the  first  place.  Hence,  the  first  place  can 
be  filled  in  9  ways.  Then  the  second  place  can  be  filled  with  any  one 
of  the  remaining  eight  letters  ;  that  is,  in  8  ways. 

In  like  manner,  the  third  place  in  7  ways,  the  fourth  place  in  G 
ways,  and  so  on  ;  and,  lastly  the  ninth  place  in  1  way. 

If  the  nine  places  are  indicated  by  Roman  numerals,  the  result 
(Rule  II.)  is  as  follows: 

I.    II.    III.  IV.  V.   VI.  VII.  VIII.  IX. 
9X8X7X6X5X4X3   X   2X1  =  362,880  ways. 

408.  Hence,  it  will  be  seen  that, 

III.    The  number  of  arrangements  or  permutations  of  n 
different  elements  or  things  taken  all  at  a  time  is 
n(n  —  l)  (n-2)  (n  —  S) 3X2X1. 


CHOICE.  321 

For,  the  first  place  may  be  filled  in  n  ways,  then  the  second  place 
in  n  —  1  ways,  then  the  third  place  in  n  —  2  ways,  and  so  on  to  the 
last  place,  which  can  be  filled  in  only  1  way. 

Hence  (Rule  II.)  the  whole  number  of  arrangements  is  the  con- 
tinued product  of  all  these  numbers, 

n{n  —  l){n-2){n  —  S) 3  X  2  X  1. 

409.  For  the  sake  of  brevity  this  product  is  written  \n, 
and  is  read  factorial  n. 

410.  It  will  also  be  evident  that, 

IV.  The  number  of  arrangements  of  n  different  elements, 
taken  t  at  a  time,  is 

n{n  —  1)  (^  —  2) to  r  factors, 

that  is,  n(n  —  1)  (/^  —  2) [n  —  (r  —  1)], 

or  n(n  —  1)  (^^  —  2) {n  —  r+1). 

For  the  first  place  can  be  filled  in  n  ways,  the  second  in  n  —  1 
ways,  the  third  place  in  n  —  2  ways,  and  the  rth  place  in  n  —  (r  —  1) 
ways. 

(1)  A  shelf  contains  4  English  books,  5  French  books,  and 

6  German  books  ;  in  liow  many  ways  can  these  books 
be  arranged  ? 

[16  =  1,307,674,368,000  ways. 

(2)  In  how  many  ways  can  these  books  be  arranged,  if  the 

books  of  each  language  are  kept  together  ? 

The  English  books  can  be  arranged  (Rule  III.)  in  [4  ways,  the 
French  books  in  [5  ways,  and  the  German  books  in  [6  ways. 
Also,  the  three  sets  can  be  arranged  in  [3  different  orders.  Hence, 
the  number  of  ways  in  which  the  whole  can  be  done  is  (Rule  II. ) 

[4XJ5X[6X|3  =  12,441,600. 


322  ALGEBRA. 

(3)  Of  the  arrangements  possible  with  the  letters  of  the 

word  Cambridge: 

(i.)    How  many  will  begin  with  a  vowel  ? 
(ii.)    How  many  will  both  begin  and  end  with  a 
vowel  ? 

In  filling  the  nine  places  of  any  arrangement  in  case  (i.),  the 
first  place  can  be  filled  in  only  3  ways,  the  other  places  in  [8 
ways. 

In  case  (ii. )  the  first  place  can  he  filled  in  3  ways,  the  last  place 
in  2  ways  (one  vowel  having  been  used),  and  the  remaining  seven 
places  in  1 7  ways. 

Hence7the  answer  to  (i.)  is  3  X  |^  =  120,960  ; 

to  (ii.)  is  3  X  2  X  [7  =  30,240. 

(4)  With  the  letters  of  the  word   Cambridge,  how  many 

arrangements  can  be  made  : 
(i.)    Each  beginning  with  the  word  Cam? 

(ii.)    Keeping  the  letters  Cam  always  together  ? 

For  case  (i.)  the  answer  is  evidently  |6  ;  since  our  only  choice 
lies  in  arranging  the  remaining  six  letters  of  the  word. 

Case  (ii.)  may  be  resolved  into  arranging  Cam  and  the  last  six 
letters,  regarded  as  seven  distinct  elements,  and  then  arranging 
the  letters  Cam. 

The  first  can  be  done  in  [7  ways,  and  the  second  in  [3  ways. 
Hence  (Rule  II.)  both  can  be  done  in  |J^  X  [3  =  30,240  ways. 

(5)  In  how  many  ways  can  the  letters  of  the  word  Cain- 

bridge  be  written : 
(i.)    Without  changing  the  place  of  any  vowel  ? 
(ii.)    Without  changing  the  order  of  any  vowel  ? 
(iii.)    Without  changing  the  relative  order  of  vowels 
and  consonants  ? 

In  (i.)  the  second,  sixth,  and  ninth  places  can  be  filled  each  in 
only  1  way  ;  the  other  places  in  [6  ways. 

Therefore,  the  whole  number  of  ways  is  |^  =  720. 

In  (ii.)  the  vowels  in  the  different  arrangements  are  always 
kept  in  the  order  a,  i,  e.  One  of  the  six  consonants  can  be  placed 
in  4  ways :  before  a,  between  a  and  i,  between  i  and  e,  and  after  e. 


CHOICE.  323 

Then  a  second  consonant  can  be  placed  in  5  ways,  a  third 
consonant  in  6  ways,  a  fourth  consonant  in  7  ways,  a  lifth  con- 
sonant in  8  ways,  and  the  last  consonant  in  9  ways.  Hence 
(Rule  II.)  the  whole  number  of  ways  is4x5x6x7x8x9 
=  60,480. 

In  (iii.)  the  vowels  can  be  arranged  in  |^  ways,  and  the  con- 
sonants in  1 6  ways.  Hence  (Rule  II.)  the  number  of  ways  is 
|_3X  1^  =  4320 

(6)  In  how  many  ways  can  4  persons,  A,  B,  C,  D,  sit  at  a 

round  table  ? 

If  the  four  places  are  not  regarded  as  relative  to  each  other, 
then  the  whole  number  of  ways  is  \_i  =  24.  But  if  the  four 
places  are  regarded  as  relative  to  each  other,  then  by  placing 
one  as  yl,  in  one  position,  and  by  arranging  the  others  in  the 
other  three  positions,  the  whole  number  of  ways  is  [3  =  G. 

(7)  In  how  many  ways  can  6  persons  form  a  ring? 

Here  relative  position  is  required.  Hence,  the  whole  number 
of  ways  is  [6  =  120. 

(8)  How  many  three-lettered  words  can  be  made  from  the 

alphabet,  no  letter  being  repeated  in  the  same  word  ? 
26  X  25  X  24  =  15,600.    Ans. 

(9)  How  many  four-lettered  words  ? 

26  X  25  X  24  X  23  =  358,800.    Ans. 

(10)  How  many  different  arrangements  can  be  made  of  the 

letters  in  the  word  et/e  ?■ 

Distinguish  the  e's  thus,  ei,  62,  and  arrange  as  follows : 
ei  y  62  ei  Ca  y  y  ei  e^ 

62  y  ei  62  eiy  y  e^  ei 

These  six  arrangements  become  three  when  the  e's  are  not 
distinguished.  That  is,  each  pair  of  arrangements  produced  by 
permuting  the  e's  is  reduced  to  a  single  arrangement. 

Hence,  the  number  of  different  arrangements  is  found  by 
dividing  |_3  by  [2  ;  that  is,  by  dividing  the  number  of  arrange- 
ments possible  when  the  letters  are  all  different  by  the  number 
of  ways  in  which  the  two  e's  can  be  permuted. 


324  ALGEBRA. 

411.  In  how  many  different  orders  can  the  letters  a,  a,  a, 
Xj  y  be  written  ? 

If  the  letters  were  all  different,  the  answer  (Rule  III.)  would  be 
1^  ;  but  the  three  a's  may  be  permuted  in  |^  =  6  ways. 

Hence,  the  [5  arrangements  may  be  divided  into  six  groups,  each 
group  constituting  but  a  single  arrangement  of  the  given  letters. 

Hence,  the  whole  number  of  given  orders  is  —  —  20, 


By  the  same  course  of  reasoning,  the  whole  number  of 
different  orders  of  a,  a,  a,  x,  x  is  found  to  be 

412.   In  like  manner  for  any  other  numbers.     Hence, 
V.    The  number  of  arrangements  of  n  elements,  of  which 
p  are  alike,  q  others  are  alike,  and  r  others  are  alike ,  is 


\p\q\r_ 


(1)    In  how  many  ways  can  the  letters  of  the  word  Missis- 
sippi be  arranged  ? 

Ill 

-  =  34,560.    Ans. 


|4[41 

(2)  In  how  many  different  orders  can  a  row  of  4  white 

balls  and  3  black  balls  be  arranged  ? 

i^  =  35.   ^n,. 

(3)  In  how  many  ways  can  4  white  balls  and  3  black  balls 

be  placed  in  a  row,  if  the  balls  are  all  different  in 

size  ? 

[7  =  5040.   Ans. 


CHOICE.  325 

413.  Ill  case  the  n  elements  to  be  arranged  are  all  differ- 
ent, but  repetitions  of  them  are  allowed,  then  in  making 
any  arrangement,  every  place  to  be  filled  can  be  filled  in  n 
ways,  since  we  may  always  repeat  an  element  already  used. 
Hence,  corresponding  to  Rules  III.  and  IV.,  the  following 
rules  apply  to  cases  in  which  repetitions  are  allowed : 

VI.  The  number  of  arrangem,ents  of  n  different  elements, 
taken  all  at  a  time,  when  repetitions  are  allowed,  is  n". 

VII.  The  number  of  arrangements  of  n  different  elements, 
taken  x  at  a  time,  when  repetitions  are  allowed,  is  viF. 

(1)  How  many  three-lettered  words  can  be  made  from  the 

alphabet,  when  repetitions  are  allowed  ? 

268  =  17,576.    Ans. 

(2)  How  many  three-lettered  words  can  be  made  from  the 

6  vowels  when  repetitions  are  allowed  ? 

68  =  216.    Ans. 

(3)  A  railway  signal  has  3  arms,  and  each  arm  may  take 

four  different  positions,  including  the  position  of 
rest.     How  many  signals  in  all  can  be  made  ? 

48-1  =  63.   Ans. 

(4)  In  the  common  system  of  notation,  how  many  numbers 

can  be  formed  consisting  of  not  more  than  5  digits  ? 

All  the  possible  numbers  may  be  regarded  as  consisting  of 
each  6  digits,  by  prefixing  zeros  to  the  numbers  consisting  of 
less  than  5  digits.     Thus,  247  may  be  written  00247. 

Hence,  every  possible  arrangement  of  5  digits  out  of  the  10 
digits  will  give  one  of  the  required  numbers  except  00000  ;  and 
the  answer  is  10^  —  1  =  99999  ;  that  is,  all  the  numbers  between 
0  and  100.000. 


326  ALGEBRA. 

(5)  With  the  digits  0,  1,  2,  3,  4,  5,  how  many  numbers 

between  1000  and  4000  can  be  formed  ? 

Here  we  have  to  fill,  in  each  possible  case,  four  places.  The 
first  place  can  be  filled  with  1,  2,  or  3,  that  is,  in  3  ways;  the 
second,  third,  and  fourth  places  each  in  6  ways.  Therefore, 
the  answer  is  3  X  6^  =  648. 

(6)  With  the  same  digits,  0,  1,  2,  S,  4,  5,  and  between  the 

same  limits,  1000  and  4000  : 
(i.)    How  many  even  numbers  can  be  formed  ? 
(ii.)    How  many  odd  numbers  can  be  formed  ? 
(iii.)    How  many  numbers  divisible  by  5  ? 

Evidently  (i.)  and  (ii.)  are  like  Ex.  (5),  except  that  the  last 
place  can  be  filled  in  only  3  ways.  In  (i.)  the  last  place  must 
be  filled  by  0,  2,  or  4 ;  in  (ii.)  the  last  place  must  be  filled  by 
1,  3,  or  5. 

Hence,  the  answer  in  each  of  these  cases  is  3x0x6x3=  324. 

In  (iii. )  the  last  digit  must  be  either  0  or  5 ;  and  the  answer 
for  this  case  is  3X6X6X2  =  216. 

Combinations. 

414.  In  how  many  ways  can  3  vowels  be  selected  from 
the  5  vowels  a,  e,  i,  o,  u  ? 

The  number  of  ways  in  which  we  can  arrange  3  vowels  out  of  5  is 
(Rule  IV.)  5  X  4  X  3  =  60. 

These  60  arrangements  might  be  obtained  by  first  forming  all  the 
possible  selections  of  the  3  vowels  out  of  5,  and  then  arranging  the  3 
vowels  in  each  selection  in  as  many  ways  as  possible. 

The  3  vowels  of  each  selection  may  be  arranged  in  [3  =  6  ways. 

Hence  (Rule  II.),  ~ 

The  number  of  selections  X  6  =  the  number  of  arrangements  =  60. 

Therefore,  the  number  of  selections  =  %^  =  10. 

415.  In  general, 

VIII.  Oict  ofn  different  elements,  the  nuniher  of  selections 
of  r  elements  is  equal  to  the  number  of  arrangements  of  n 
elements  divided  by  [r. 


CHOICE.  327 

For,  let  8  —  the  number  of  ways  of  selecting  r  elements  out  of  n 
elements.  Then  the  r  elements  thus  selected  may  be  arranged 
(Rule  III.)  in  \r  different  ways.  Therefore  (Rule  I.)  s  X  |r=  the 
number  of  arrangements  of  n  elements  taken  r  at  a  time. 

_  the  number  of  arrangements 

(By  Rule  IV.)    The  numerator  of  this  fraction  is  equal  to 

n  {n  -  1)  (n  -  2) [n-{r-  1)]. 

_  n  (n  —  1)  (n  —  2) (n  —  r  +  1) 

■'■'-      ■  \r 

If  both  terms  of  this  fraction  are  multiplied  by  |  n  —  r, 
the  result  is 

If  n  —  r  =  p,  then  n  =  p+  r,  and  this  formula  may  be  written 
_\p_±_r 

416.  The  value  of  this  fraction  is  not  altered  if  p  and  r 
are  interchanged.     Hence, 

IX.  Out  o/  p  +  r  different  elements,  the  number  of  ways 
in  which  p  elements  can  he  selected  is  the  same  as  the  number 
of  ways  in  which  r  elements  can  be  selected. 

Thus,  out  of  8  elements,  3  elements  can  be  selected  in  the  same 
number  of  ways  as  5  elements ;  namely, 

IJ        8.7.6       ^^ 
_  =  — =  56ways. 

(1)    Out  of  20  consonants,  in  how  many  ways  can  18  be 
selected  ? 

The  18  can  be  selected  in  the  same  number  of  ways  as  2  ;  and 
the  number  of  ways  in  which  2  can  be  selected  (Rule  VIII.)  is 

20  X  19       ,._       .  , 

— =  190.    Ans.  * 


328  ALGEBRA. 

(2)  In  how  many  ways  can  the  same  choice  be  made  so  as 

always  to  include  the  letter  B  ? 

Taking  B  lirst  we  must  then  select  17  out  of  the  remaining  19 
consonants.     This  can  be  done  in 

1,9  X  18      ,^, 

^^-^ —  =  171  ways. 

(3)  In  how  many  ways  can  the  same  choice  be  made  so  as 

to  include  B  and  not  to  include  C  ? 

Taking  B  first,  we  have  then  to  choose  17  out  of  18,  C  being 
excluded.     This  can  be  done  in  18  ways. 

(4)  A  society  consists  of  50  members,  10  of  whom  are 

physicians.  In  how  many  ways  can  a  committee  of 
6  members  be  selected  so  as  to  include  1  physician  ? 
The  6  members  not  physicians  can  be  selected  in 

[40 


[5135 


ways, 


and  the  1  physician  in  10  ways.    Hence,  the  6  can  be  selected  in 

|40 
10  ><  I  ^  ,or  different  ways. 

(5)    In  how  many  ways  can  a  committee  of  6  members  be 
selected  so  as  to  include  at  least  1  physician  ? 

Six  members  can  be  selected  from  the  whole  society  in 

|50 

Six  members  can  be  selected  from  the  whole  society,  so  as  to 
include  no  physician^  by  choosing  them  all  from  the  40  members 
who  are  not  physicians,  and  this  can  be  done  in 

[6134  ^^y«- 

[50  [40 

Hence,  .„  ,, .  —  ,„  ,,,  =  the  number  of  ways  of  select- 

I o  1 44        |0  1  'j4 

ing  the  committee  so  as  to  include  at  least  one  physician. 


CHOICE.  329 

(6)  Out  of  20  Eepublicans'and  6  Democrats,  what  choice 

is  there  of  appointing  a  committee  consisting  of  3 
Kepublicans  and  2  Democrats  ? 

20  X  19  X  18 

The  Republicans  can  be  selected  in  — —  =  1140  ways; 

0X5  lX2Xo 

and  the  Democrats  in  -         =15  ways.     Hence,  the  whole  com- 
1  X  ^ 

mittee  can  be  appointed  in  1140  x  15  =  17,100  ways. 

(7)  From  20  Eepublicans  and  6  Democrats,  in  how  many- 

ways  may  5  different  offices  be  filled,  three  of  which 
must  be  filled  by  Republicans,  and  the  other  two  by- 
Democrats  ? 

The  first  three  offices  can  be  assigned  to  3  Republicans  in 
20  X  10  X  18  =  C840  ways  (Rule  II.) ;  and  the  other  two  offices 
can  be  assigned  to  2  Democrats  in  6  X  5  =  30  ways. 

There  is,  then,  a  choice  of  6840  x  30  =  206,200  different 
ways. 

(8)  Out  of  20  consonants  and  6  vowels,  in  how  many  ways 

can  we  make  a  word  consisting  of  3  different  conso- 
nants and  2  different  vowels  ? 

20  X  19  X  18 

Three  consonants  can  be  selected  in  — -—  =  1140 

6x5  lX2X,j 

ways,  and  2  vowels  in  7— „  =  15  ways.     Hence  (Rule  I.)  the  6 

letters  can  be  selected  in  1140  X  15  =  17,100  ways. 

When  they  have  been  so  selected,  they  can  be  arranged  (Rule 
III.)  in  1^  =  120  different  orders.  Hence,  there  are  17,100  X 
120  =  2,052,000  different  ways  of  making  the  word. 

(9)  How  many  words  of  2  consonants  and  1  vowel  can  be 

formed  from  6  consonants  and  3  vowels,  the  vowel 
being  the  middle  letter  of  each  word  ? 

The  two  consonants  can  be  selected  in  15  ways;  the  vowel  in 
3  ways.  Each  combination  of  the  two  consonants  and  1  vowel 
can  be  arranged  in2X  IX  1  =  2  ways.  Hence,  the  number  of 
words  that  can  be  formed  is  15  X  3  X  2  =  90. 


330  ALGEBRA. 

(10)  How  many  words  of  3  consonants  and  3  vowels  can 
be  formed  from  the  alphabet,  if  one  of  the  vowels 
is  to  be  always  a  ? 

T    .  ^  •    20  X  19  X  18     ,,,_ 
The  consonants  can  be  selected  m     ,  ^  ^  ^  „     =1140  ways, 
5x4  ix^xd 

and  the  vowels  in  i— — ^  =  10  ways, 
i  X  z 

Then  the  six  letters  of  each  combination  can  be  arranged  in 

[6  :=  720  ways.    Hence  the  number  of  words  that  can  be  formed 

is  1140  X  10  X  720  =  8,208,000. 

417.  To  find  for  what  value  of  r  the  number  of  selections 
of  n  things,  taken  t  at  a  time,  is  greatest. 

^     ,         ,  n(n  —  l){n  —  2) (n  —  r+l) 

Theformnla     s=   -'i^l,^t..,   

n      n  —  I      n  —  2         n  —  r+l 

may  be  written    s  =  -  X  — - —  X  — r—  

i  ^  o  r 

The  numerators  of  the  factors  on  the  right  side  of  this  equation 
begin  with  n,  and  form  a  descending  series  with  the  common  differ- 
ence 1 ;  and  the  denominators  begin  with  1,  and  form  an  ascending 
series  with  the  common  difference  1.  Therefore,  from  some  point  in 
the  series,  these  factors  become  less  than  1.  Hence,  the  maximum 
product  is  reached  when  that  product  includes  all  the  factors  greater 
than  1. 

When  n  is  an  odd  number,  the  numerator  and  the  denominator 
of  each  factor  will  be  alternately  both  odd  and  both  even ;  so  that 
the  factor  greater  than  1,  but  nearest  to  1,  will  be  the  factor  whose 
numerator  exceeds  the  denominator  by  2.  Hence,  in  this  case,  r  must 
have  such  a  value  that 

n  —  1 

n  —  r+l  =  r+2,     or    r=  • 

When  n  is  an  even  number,  the  numerator  of  the  first  factor  will 

be  even  and  the  denominator  odd  ;  the  numerator  of  the  second  factor 

will  be  odd  and  the  denominator  even ;  and  so  on,  alternately ;  so 

that  the  factor  greater  than  1,  but  nearest  to  1,  will  be  the  factor 

whose  numerator  exceeds  the  denominator  by  1.    Hence,  in  this  case, 

r  must  have  such  a  value  that 

1111  '^ 

n  — r+l  =  r  +  l,     or     '>'  =  -' 


CHOICE.  331 

(1)  What  value  of  r  will  give  the  greatest  number  of  selec- 

tions out  of  7  things  ? 

Here  n  is  odd,  and  r  —  — - —  =  — - —  =  3. 

„         .    ,,  7X6X5X4      „, 

Ifr  =  4,the„  «=1><2XSX4  =  «^- 

So  that,  when  the  number  of  things  is  odd,  there  will  be 
two  equal  numbers  of  selections  ;  namely,  ichen  the  number 
of  things  taken  together  is  just  under  and  just  over  one-half 
of  the  whole  number. 

(2)  What  value  of  r  will   give  the  greatest  number  of 

selections  out  of  8  things  ? 

Here  n  is  even,  and  r  =  -  =  -  =  4. 

8x7x6x5      ^^      , 
•••*=1X2X3X4^'^-    ^^- 

So  that,  when  the  number  of  things  is  even,  the  number 
of  selections  will  be  greatest  when  one-half  of  the  whole 
are  taken  together. 

418.    It  may  be  shown  that, 

X.  The  number  of  ways  in  which  x  -|-  y  different  elements 
can  be  divided  into  two  classes,  so  that  one  shall  contain  x 
and  the  other  y  elements,  is  equal  to  the  nimiber  of  ways  in 
which  either  x  elements  or  y  elements  may  be  selected  from 
the  X  -j-  y  elements  ;  or 

For  each  division  oix  -{■  y  elements  into  two  classes,  one  consisting 
of  X  elements,  the  otlier  of  y  elements,  is  evidently  effected  by  making 
a  selection  of  x  elements  from  x-\-  y  elements,  and  leaving  y  elements 
not  selected. 


332  ALGEBRA. 

In  like  manner  the  number  of  ways  in  which  x-\-  i/-\-z 
different  elements  can  be  divided  into  3  classes,  containing 
X,  y,  and  z  elements  respectively,  is 

\x-\-i/  +  z 


\x\y\z 


and  so  on. 


(1)  In  how  many  ways  can  18  men  be  divided  into  2  classes 

of  6  and  12  ? 

mi 

(2)  In  how  many  ways  can  18  men  be  divided  into  2  groups 

of  9  each  ? 

According  to  the  rule,  the  answer  would  be 

|_9|_9 

The  two  groups,  considered  as  groups,  have  no  distinction ; 
therefore,  permuting  them  gives  no  new  arrangement,  and  the 
true  result  is  obtained  by  dividing  the  preceding  by  [2,  and  is 

jig 

[2|^|_9 

If  any  condition  is  added  that  shall  make  the  two  groups 
different,  as,  if  one  group  wear  red  badges  and  the  other  blue, 
then  the  answer  will  be 

|9[9 

419.  Whenever  the  groups  are  indifferent,  Kule  IX.  gives 
each  arrangement  repeated  as  many  times  as  the  groups  can 
be  permuted  one  with  another  ;  that  is,  [2  when  there  are 
2  groups,  1 3  when  there  are  3  groups,  and  so  on. 

Hence,  the  result  found  by  the  rule  must  be  divided  by 
|2,  |3,  etc.,  in  order  to  obtain  the  true  result. 


CHOICE.  333 

(1)  In  how  many  ways  can  the  52  cards  in  a  pack  be 

divided  among  4  players,  each  to  have  13  ? 

Here  the  assignment  of  each  group  to  a  different  player  makes 
the  groups  different ;  and  the  answer  is 

|_52 
|T3J13J13]13' 

(2)  In  how  many  ways  can  the  52  cards  of  a  pack  be 

divided  into  4  piles  ? 

Here  the  groups  are  indifferent,  and  the  answer  is 

[4|J^|_13|^|J^' 

(3)  A  boat's  crew  consists  of  8  men,  of  whom  2  can  row 

only  on  the  stroke  side  of  the  boat,  and  3  can  row 
only  on  the  bow  side.  In  how  many  ways  can  the 
crew  be  arranged  ? 

There  are  left  3  men  who  can  row  on  either  side ;  2  of  these 
must  row  on  the  stroke  side,  and  one  on  the  bow  side. 

The  number  of  ways  in  which  these  three  can  be  divided  is 

^=3  ways. 

When  the  stroke  side  is  completed,  the  4  men  can  be  arranged 
in  |_4  ways ;  likewise,  the  4  men  of  the  bow  side  can  be  arranged 
in|_4  ways.  Hence  (Rule  II.)  the  arrangement  can  be  made  in 
3X[4[4=  1728  ways. 

(4)  In  how  many  ways   can  10   copies  of  Homer,  6  of 

Virgil,  and  4  of  Horace  be  given  to  20  boys,  so 
that  each  boy  may  receive  a  book  ? 

The  boys  have  to  form  themselves  into  a  group  of  10  for 
Homer,  of  6  for  Virgil,  of  4  for  Horace.     This  can  be  done  in 

1 20 


[lOl^lJ 


different  ways. 


334  ALGEBRA. 

(5)  In  how  many  ways  can  3  copies  of   one  book,  2  of 

another,  and  1  of  a  third  be  given  to  a  class  of  12 
boys,  so  that  no  boy  shall  receive  more  than  1  book  ? 

In  every  possible  way  of  assigning  the  books,  6  boys  receive 
them.     These  6  may  be  selected  from  the  whole  12  (Rule  VIII.) 

in  [12 

,-^ways. 

When  thus  selected,  the  books  may  be  assigned  to  them 
(Rule  V.)  in  |6 

Hence,  the  whole  number  of  ways  of  giving  the  books  is 

[6  1_6  ^  |3  [2  1_1 ' 

(6)  In  how  many  ways  can  the  same  books  be  given  to 

the  12  boys,  so  that  no  boy  shall  receive  more  than 
1  copy  of  any  book  ? 

In  allotting  the  three  copies  of  the  first  book,  the  boys  are  to 

be  separated  into  two  groups  of  3  and  9 ;  and  the  group  of  3 

will  in  each  case  receive  the  books. 

IJ2 
This  can  be  done  in  -r—  ways. 

Likewise  the  two  copies  of  the  second  book  may  be  given  in 

|_12 

The  single  copy  of  the  third  book  can  be  given  in  12  ways. 
Hence  (Rule  II.)  the  books  may  be  given  in 
|_12  IJ^ 

ll|9  ""  |T^  ""  ^^  ^^^""'"'^  '^^^''• 

(7)  In  how  many  ways  can  2  letters  be  selected  from  a,  b, 

c,  d,  e,  /,  if  the  letters  may  be  repeated  in  making 
the  selection  ? 

Without  repetitions,  two  letters  can  be  selected  from  6  in  15  ways. 
With  repetitions,  as  aa,  etc.,  6  selections  can  be  made.  Hence,  there 
are  15  +  0  =  21  different  selections. 


CHOICE.  335 

(8)    In  how  many  ways  can  selections  of  3  letters  be  made 
from  a,  b,  c,  d,  e,  f,  if  letters  may  be  repeated  ? 

These  selections  will  be  of  3  kinds : 
(i.)  All  three  letters  different, 
(ii.)  Two  letters  alike,  the  third  different, 
(iii.)  All  three  letters  alike. 
(By  Rule  VIII.)  (i.)  gives  20  ways. 

(ii.)  gives  6  X  5  =  30 ;  for  we  can  choose  2  alike  in  6  ways, 
and  then  join  a  different  letter  to  each  pair  in  5  ways, 
(iii.)  gives  evidently  6  ways. 
Hence,  there  are  in  all  20  +  30  -|-  G  =  56  different  selections. 

420.   This  may  also  be  shown  as  follows  : 

By  adding  the  6  letters,  a,  h,  c,  d,  e,  /,  to  each  selection  of 
the  kind  required,  they  will  become  selections  of  6  +  3  =  9 
letters  out  of  6,  in  which  selections  each  letter  occurs  at 
least  once^  and  in  which,  therefore,  there  must  be  exactly 
3  repetitions.  These  repetitions  may  be  all  of  the  same 
letter,  or  divided  among  the  different  letters. 

Any  one  of  these  selections  may  be  made  by  writing  9 
places,  and  then  filling  them  in  alphabetical  order,  taking 
care  to  make  exactly  three  repetitions  in  passing  from  one 
end  of  the  row  to  the  other. 

Thus  : 

12  3  4  5  6  7  8 

aaaahcdef 
ahcccdeef 
abccddeef 

By  striking  out  the  6  letters,  a,  h,  c,  d,  e,  f,  each  once, 
there  is  left  in  the  first  row  aaa,  in  the  second  row  cce,  in 
the  third  row  cde ;  tliat  is,  three  of  the  required  selections. 

Now,  in  filling  the  9  places  for  each  row,  9  —  1  or  8  steps 
must  be  made,  of  whicli  exactly  3  are  repetitions,  and  each 
of  the  other  ^  is  a  chano:e  to  a  different  letter. 


336  ALGEBRA. 

The  3  repetitions  may  be  chosen  from  the  8  steps 
(Rule  VIII.)  in 

8X7X6       ^^ 

This,  then,  is  the  number  of  ways  in  which  3  letters  can 
be  selected  out  of  6  letters,  when  i^epetitions  are  allowed. 

In  other  words,  3  elements  can  be  selected  from  6 
elements,  when  repetitions  are  allowed,  in  as  many  ways 
as  3  elements  can  be  selected  from  6  +  3  —  1,  or  8  elements 
without  repetitions. 

421.   Hence  the  general  rule  : 

XI.  The  number  of  ways  of  selecting  r  elements  from  n 
different  elements,  when  repetitions  are  allowed,  is  the  same 
as  the  number  of  ways  of  selecting  r  elements  from  n  -f"  r  —  1 
elements  without  repetitions. 

And  this  number  of  ways  is  (Rule  VIII.). 

\n-\-r—\  _  n{n-\-V){n-\-2) {n-\-r  —  V) 

\r\n  —  1  \r 

(1)  In  how  many  ways  can  4  elements  be  selected  from  n 

elements,  when  repetitions  are  allowed  ? 

n  (n  +  1)  (n  +  2)  (n  +  3)     ^^^ 

li 

(2)  How  many  dominoes  are  there  in  a  set  mumbered  from 

double  blank  to  double  nine  ? 

Each  domino  is  made  by  selecting  two  numbers  out  of  the  ten 
digits,  and  repetitions  are  included  ;  that  is,  the  two  numbers  on 
a  domino  may  be  the  same. 

Hence,  the  number  of  dominoes  is  equal  to  the  number  of 
selections  of  2  from  10  +  2  —  1,  or  11,  without  repetitions. 

10X11       ^^       . 

T>r2-  =  ^^-  ^^^- 


CHOICE.  337 

(3)    In  how  many  ways  can  4  glasses  be  filled  with  5  kinds 
of  wine,  without  mixing  ? 

The  numT3er  is  equal  to  the  number  of  ways  in  which  4  things 

can  be  selected  from  5  +  4  —  1,  or  8  things,  without  repetitions. 

5  X  0  X  7  X 


1X2X3X4 


=  70.    Ans. 


(4)    In  how  many  ways  can  six  rugs  be  selected  at  a  shop 
where  2  kinds  of  rugs  are  sold  ? 

The  number  is  equal  to  the  number  of  ways  of  selecting  6 
from  2  +  6—1  =  7,  without  repetitions. 

2  X  3  X  4  X  5  X  6J<J7  _ 

rx  2  X  3  X  4  X  5  X  G  ~    ■ 
If  a  and  b  represent  the  two  kinds  of  rugs,  the  7  ways  are  as 

follows : 

a  a  a  a  a  a  a  a  abb  h 

a  a  a  a  a  b  a  abb  b  b 

a  a  a  a  b  b  abb  b  b  b 

bbbbbb 

422.   It  may  be  shown  that, 

XII.  The  number  of  ways  in  which  a  selection  (of  some, 
or  all)  can  he  made  from  n  different  things  is  2"  —  1. 

For  each  thing  can  be  either  taken  or  left,  that  is,  can  be  disposed 
of  in  two  ways. 

There  are  n  things ;  hence  (Rule  II.)  they  can  all  be  disposed  of  in 
2"  ways.  But,  among  these  ways  is  included  the  case  in  which  all  are 
rejected  ;  and  this  case  is  inadmissable. 

Hence,  the  number  of  ways  of  making  the  selection  is  2"  —  1. 

(1)  In  a  shop  window  20  different  articles  are  exposed  for 

sale.     What  choice  has  a  purchaser  ? 
220  _  1  =  1,048,575.    Ans. 

(2)  How  many  different  amounts   can  be  weighed  with 

1  lb.,  2  lb.,  4  lb.,  8  lb.,  and  16  lb.  weights? 

25  —  1  =  31.    Ans. 

(Let  the  student  write  out  the  31  weights.) 


338 


ALGEBRA. 


423.    It  may  be  shown  that, 

XIII.    The  whole  number  of  ways  in  which  a  selection  can 

be  made  from  p  -|-  q  -f"  r things,  of  which  p  are  alike,  q  are 

alike,  r  are  alike,  etc.,  is  (p  +  1)  (^  +  1)  (r  +  1) —  1. 

For  the  set  of  -p  things  may  be  disposed  of  in  _p  +  1  ways,  since 
none  of  them  may  be  taken,  or  1,  2,  3,  ,  or  _p,  may  be  taken. 

In  like  manner,  the  g  things  may  be  disposed  of  in  g  +  1  ways ; 
the  r  things  in  r  +  1  ;  and  so  on. 

Hence  (Rule  II.),  all  the  things  may  be  disposed  of  in  (p  +  1) 
(g  +  1)  (r  +  1) ways. 

But  the  case  in  which  all  the  things  are  rejected  is  inadmissable  ; 
hence,  the  whole  number  of  ways  is  {jy  +  1)  (g  +  1)  (r  +  1) —  1. 

(1)  In  how  many  ways  can  2  boys  divide  between  them 

10  oranges   all  alike,  15  apples   all  alike,  and  20 
peaches  all  alike  ? 

Here,  the  case  in  which  the  first  boy  takes  none,  and  the  case 
in  which  the  second  boy  takes  none,  must  be  rejected. 

Therefore,  the  answer  is  1  less  than  the  result,  according  to 
Rule  XIII.     11  X  IG  X  21  —  2  =  8694.    ^ns. 

(2)  If  there  are  iti  kinds  of  things,  and  n  things  of  each 

kind,  in  how  many  ways  can  a  selection  be  made  ? 

In  this  case  p,  g,  r,  etc.,  are  all  equal,  and  each  is  equal  to  n. 
Hence  the  result  is  (n  +  1)"'  —  1. 

(3)  If  there  are  m  kinds  of  things,  and  1  thing  of  the  first 

kind,  2  of  the  second,  3  of  the  third,  and  so  on,  in 
.  how  many  ways  can  a  selection  be  made  ? 
|m+  1  —  1.   Ans. 

Exercise  119. 

1.  How  many  different  permutations  can  be  made  of  the 

letters  in  the  word  Ecclesiastical,  taken  all  together? 

2.  Of  all  the  numbers  that  can  be  formed  with  four  of  the 

digits  5,  6,  7,  8,  9,  how  many  will  begin  with  h^  ? 


CHoici^.  339 

3.  If  the  number  of  permutations  of  n  things,  taken  4 

together,  is  equal  to  12  times  the  permutations  of 
n  things,  taken  2  together,  find  n. 

4.  With  3  consonants  and  2  vowels,  how  many  words  of  3 

letters  can  be  formed,  beginning  and  ending  with  a 
consonant,  and  having  a  vowel  for  the  middle  letter  ? 

5.  Out  of  20  men,  in  how  many  different  ways  can  four 

be  chosen  to  be  on  guard  ?  In  how  many  of  these 
would  one  particular  man  be  taken,  and  from  how 
many  would  he  be  left  out  ? 

6.  Of  12  books  of  the  same  size,  a  shelf  will  liold  5. 

How  many  different  arrangements  on  the  shelf  may 
be  made  ? 

7.  Of  8  men  forming  a  boat's  crew,  one  is  selected  as 

stroke.  How  many  arrangements  of  the  rest  are 
possible  ?  When  the  4  wlio  row  on  each  side  are  de- 
cided on,  how  many  arrangements  are  still  possible  ? 

8.  How  many  signals  may  be  made  with  6  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or 
any  number  at  a  time  ? 

9.  How  many  signals  may  be  made  with  8  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or 
any  number  at  a  time  one  above  another  ? 

10.  How  many  different  signals  can  be  made  with  10  flags, 

of  which  3  are  white,  2  red,  and  the  rest  blue, 
always  hoisted  all  togetlier  and  one  above  another  ? 

11.  How  many  signals  can  be  made  with  7  flags,  of  which 

2  are  red,  1  white,  3  blue,  and  1  yellow,  always  dis- 
played all  together  and  one  above  another  ? 

12.  In  how  many  different  ways  may  the  8  men  serving  a 

field-gun  be  arranged,  so  that  the  same  man  may 
always  lay  the  gun  ? 


340  ALGEiJKA. 

13.  Find  the  number  of  signals  which  can  be  made  with 

4  lights  of  different  colors  when  displayed  any 
number  at  a  time,  arranged  above  one  another,  side 
by  side,  or  diagonally. 

14.  From  10  soldiers  and  8  sailors,  how  many  different 

parties  of  3  soldiers  and  3  sailors  can  be  formed  ? 

15.  How  many  signals   can  be  made  with  3  blue  and  2 

white  flags,  which  can  be  displayed  either  singly, 
or  any  number  at  a  time  one  above  another  ? 

16.  In  how  many  ways  can  a  party  of  6  take  their  places 

at  a  round  table  ? 

17.  Out  of  12  Democrats  and  16  Kepublicans,  how  many 

■  different  committees  can  be  formed,  each  consisting 
of  3  Democrats  aud  4  Republicans  ? 

18.  From  12  soldiers  and  8  sailors,  how  many  different 

parties  of  3  soldiers  and  2  sailors  can  be  formed  ? 

19.  Find  the  number  of  combinations  of  100  things,  97 

together. 

20.  With  20  consonants  and  5  vowels,  how  many  different 

words  can  be  formed  consisting  of  3  different  conso- 
nants and  2  different  vowels,  any  arrangement  of 
letters  being  considered  a  word  ? 

21.  Of  30  things,  how  many  must  be  taken  together  in 

order  that  having  that  number  for  selection,  there 
may  be  the  greatest  possible  variety  of  choice  ? 

22.  There  are  m  things  of  one  kind  and  n  of  another ;  how 

many  different  sets  can  be  made  containing  r  of  the 
first  and  s  of  the  second  ? 

23.  In  how  many  ways  may  10  persons  be   seated  at  a 

round  table,  so  that  in  no  two  of  the  arrangements 
may  every  one  have  the  same  neighbors  ? 


CHOICE.  341 

24.  The  number  of  combinations  of  n  things,  taken  r  to- 

gether, is  3  times  the  number  taken  r  —  1  together, 
and  half  the  number  taken  r-\-l  together.  Find  n 
and  r. 

25.  In  how  many  ways  may  12  things  be  divided  into  3 

sets  of  4  ? 

26.  How  many  words  of  6  letters  may  be  formed  of  3 

vowels  and  3  consonants,  the  vowels  always  having 
the  even  places? 

27.  From  a  company  of  90  men,  20  are  detached  for  mount- 

ing guard  each  day.  How  long  will  it  be  before  the 
same  20  men  are  on  guard  together,  supposing  the 
men  to  be  changed  as  much  as  possible ;  and  how 
many  times  will  each  man  have  been  on  guard  ? 

28.  Supposing  that  a  man  can  place  himself  in  3  distinct 

attitudes,  how  many  signals  can  be  made  by  4  men 
placed  side  by  side  ? 

29.  How  many  different  arrangements  may  be  made  of  11 

cricketers,  supposing  the  same  two  always  to  bowl  ? 

30.  Five  flags  of  different  colors   can  be  hoisted  either 

singly,  or  any  number  at  a  time  one  above  another. 
How  many  different  signals  can  be  made  with  them  ? 

31.  How  many  signals  can  be  made  with  5  lights  of  differ- 

ent colors,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  side  by  side,  or  one  above 
another  ? 

32.  The  number  of  permutations  of  n  things,  3  at  a  time, 

is  6  times  the  number  of  combinations,  4  at  a  time. 
Find  n. 

33.  At  a  game  of  cards,  3  being  dealt  to  each  person,  any 

one  can  have  425  times  as  many  hands  as  there  are 
cards  in  the  pack.     How  many  cards  are  there  ? 


342  ALGEBKA. 


Binomial  Theorem. 

424.    By  performing  the  indicated  multiplication 

(x  -\-  a)  {x-\-h)^  x^  -\- {(t -\- h)  X -\-  ah. 

Multiply  each  side  by  x  -\-  c,  and  the  result  is 

{x  -\-a){x-{-  h)  (x + c)=^x^-\-{(i  -\-h)x'^       -\-  ahx 

-f-  cx^       -\-  (ac-\-hc)x-\-ahc 

=x^-\-{ii-]rh-\-G)x^-\-{ab-\-ac-\-bc)x-\-abc 

rrom  this  result  certain  laws  are  to  be  observed  : 

I.  The  number  of  terms  is  one  more  than  the  number  of  factors  on 
the  left  side. 

II.  The  exponent  of  x  in  the  first  term  is  the  same  as  the  number 
of  factors,  and  decreases  by  1  in  each  succeeding  term. 

III.  The  coefficient  of  x  in  the  first  term  is  unity  ; 
in  the  second  term,  the  sum  of  a^h^  c, 

in  the  third  term,  the  sum  of  the  products  two  and  two,  a6,  ac^  he ; 
and  the  fourth  term  is  the  product  ahc. 

Do  the  same  laws  hold,  whatever  be  the  number  of  factors  ? 

Suppose  that  these  laws  hold  for  r  factors,  so  that 

{x  +a){x-\-  h) (x  +  m)=  x''  +  piX'--^+  p2X''-^  +  PsX''-"  + +Pr 

where  pi  stands  for  a+  b  + +  ?/i,  the  sum  of  the  second  terms  ; 

P2  stands  for  ab  +  ac-\- ,  the  sum  of  the  products,  two  and 

two ; 
Ps  stands  for  ahc  +  abd+ ,  the  sum  of  the  products,  three 

and  three  ; 
Pr  stands  for  abed ,  the  product  of  the  r  letters. 

Multiply  by  another  factor  {x  +  n),  and  the  product  of  the  r  +  1 
factors  is 

x'"+i+     PiX'-    +    PiX"—^  +    psx"--^  + 4-      PrX 

+     nx>'      4-     piiix''-^      +     p2nx''-^      + +      p,.-inx    +Prn 

=x'+i+(pi+n)x'+(p2+i>iw)x'— H(i93+P2w)iC'— 2+ +{p,.-^,._in)x-\-prn 

Here  the  laws  I.  and  II.  evidently  hold ;  and  as  to  the  coefficients 
Pi-\-  n=  a  +  b  + +  rn  +  n,  the  sum  of  r  +  1  letters  ; 


BINOMIAL    THEOREM.  343 

P2  +  Pin  =  (ab  +  ac  + )  +  (an  +  bn  + +  mn),  the  sum  of  the 

products,  two  and  two  ; 

Ps  +  P^n  =  {abc  +  abd  + )  +  {abn  +  acn  + ),  the  sum  of  the 

products,  three  and  three. 


PrU  =  abc mn,  the  product  of  the  r  +  1  letters.     Hence, 

The  laws  hold  for  r  +  1  factors  if  they  hold  for  r  factors. 

But  they  have  been  shown  to  hold  for  three  factors,  therefore  they 

hold  for  four  factors,  and  therefore  for  five  factors;  and  so  on  for  any 

number  of  factors.* 

425.    When  the  factors  in  the  preceding  proof  are  all 

equal,  so  that  b,  c,  d,  n,  are  each  equal  to  a,  the  left 

side  of  the  equation  becomes 

(x  4-  a)  (x  +a) taken  n  times  ;  that  is,  (x  +  a)". 

On  the  right  side. 

Pi  =  a  +  a-\-  =  a  taken  n  times  =  na  ; 

P2  =  aa  +  aa+  =  a-^  taken  as  many  times  as  there  is  a  choice 

of  2  letters  from  n  letters, 

that  is,  P2  =  ^\^~c}^  a"^ ;  §  415. 

^3  =  aaa  +  aaa  + =  a^  taken  as  many  times  as  there  is  a  choice 

of  3  letters  from  n  letters, 

4.x.  ^  •  n(n  — l)(n  — 2)    -  ..^. 

that  IS,  Pa  =     \  ^  ,  ^  ., — -  a^.  §  415. 

1  X  z  X  o 


Prn  =  ax  a  X  a =  a  taken  n  times  as  a  factor  =  a". 

.-.  (X  +  ay  =  x"  +  nax"-^  +  ^  ["'~^'^  a^  x"--' 

1  X  ^ 
1  X  J  X  o 

426.  The  expression  on  the  right  side  is  called  the 
expansion  of  (ic +  <■/)". 

If  a  and  x  are  interchanged,  the  expansion  will  proceed 
by  ascending  powers  of  x,  as  follows  : 


a  +  x)«  =  a«  +  na^-^  x  +  ^  |^^  ^^  a«-2x2  + +  nax^-'^  +  x". 

*  A  proof  of  this  kind  is  called  mathematical  induction. 


344  ALGEBRA. 


If  a=  1,  then, 


(1  +  x)«  =^  1  +  nx  +  ^  i\  J^  x2  + +  nxn-i  +  x«. 

1  X  ^ 

If  X  is  negative,  the  odd  powers  of  x  will  be  negative  and  the  even 

powers  positive. 

^      ,  n  (n  — 1) 

(a  —  x)'*  =  a"  —  iia«-i  x  H ^      ^  a»- 

i  X  ^ 


-2x2 


_  n(n-l)(n-2) 

1X2X3       «    ^  ^ 


427.  It  will  be  observed  that  the  last  factor  in  the  de- 
nominator of  the  coefficient  is  one  less  than  the  number  of 
the  term;  and  is  the  same  as  the  expo7ient  of  the  second  letter ; 
also,  that  the  last  factor  of  the  7iumerator  of  the  coefficient 
is  found  by  subtracting  the  last  factor  in  the  denominator 
from  n-\-l,  and  that  the  exponent  of  the  first  letter  is  found 
by  subtracting  the  exponent  of  the  second  letter  from  n. 
So  that, 

The  rth  (or  general)  term  in  the  expansion  of  {a  +  x)«  is 


yi(n-l) (n  -  r  +  2) 

1  X  2.....(r-l) 

Thus,  the  third  term  of  (a  +  x)2'^  is 

20  X  19 


^n-r+\xr 


1X2 


ai8x2  =  190  a28x2. 


428.    The  coefficient  of  the  rth  term  from  the  beginning 
is  equal  to  the  coefficient  of  the  rth  term  from  the  end. 

For,  the  coefficient  of  the  rth  term  from  the  beginning  is 

n{n—  1) (w  — r  +  2)  n{n—  1) {n  —  r-\-  2)  _ 

1  X2 (r-1)        '    °^  |r-l  ' 

and  this  becomes,  when  both  terms  are  multiplied  by  jn  — r  +  1 , 

\n^ 

|r  —  1  |n  —  r  +  1 


BINOMIAL    THEOREM.  345 

The  coefficient  of  the  rth  term  from  the  end,  which  is  the  {n—r+2)th 
term  from  the  beginning,  is 

n  (n  —  1) r 

I  ?i  -  r  +  1  "  ' 

and  this  also  becomes,  when  both  terms  are  multiplied  by  |r  —  1 , 

\n 
\r-l  |n-r  +  l' 

429.  It  will  be  evident  from  §  417,  that  in  the  expansion 
of  (a-{-xy,  the  middle  term  will  have  the  greatest  coeffi- 
cient when  71  is  even ;  and  when  n  is  odd  the  two  middle 
terms  will  have  equal  coefficients,  and  these  will  be  the 
greatest. 


Exercise  120. 
Expand : 

1.    (l  +  2xy.  3.    (2x  —  3i/y. 


2.  (x—sy.  4.  (2—xy. 

7.  Find  the  fourth  term  of  (2x  —  oi/y-. 

8.  Find  the  seventh  term  of  (  |  + 1)   * 

9.  Find  the  twelfth  term  of  (a-  —  axy^ 
10.    Find  the  eighth  term  of  (5xy  —  2xf)^ 


11.    Find  the  middle  term 


«'(M)' 


12.    Find  the  middle  term  of  '  ^      ^ 


13.    Find  the  two  middle  terms  of 


346  ALGEBRA. 

14.  Find  the  rth  term  of  (2ft^-^)^ 

15.  Find  the  rth  term  from  the  end  of  (2a-\-xy. 

16.  Find  the  {T-\-^)th  term  of  (a-\-xy. 

17.  Find  the  middle  term  of  (o^-}-^)^". 

18.  Expand  (2a-\-xy^,  and  find  the  sum  of  the  terms  if 

a  =  l,  x=^  —  2. 

When  the  Exponent  is  Fractional  or  Negative. 

430.   When  n  is  a  whole  number,  §§  424-426. 

XH.    .      s         -.    .  ,   oi(n  —  1)    „   ,   11  (n  —  V)  (n  —  2)    „  , 

P 

Suppose  ?z  to  be  a  positive  fraction  -•     We  may  take 

(lJ^x)\  =  l-\-Ax-\- 

By  raising  both  sides  to  the  qth  power,  we  have 

{i^xyr=(ij^Ax-\- y,- 

or  (l-\-px-\- )=^l-[-qAx-\- 

P 
Hence,  qA=p,  and  therefore  A=  — 

That  is,  the  numerical  coefficient  of  x  is  the  same  as  the 
exponent  of  the  binomial,  when  the  exponent  is  a  positive 
fraction. 

Supi)ose  n  is  negative  =  —  m. 
Then     (1 +a-)"  =  (l -f  x)-"  ^ 


(1+^)- 

=  T— -^-T §§  424-426. 

=  1  —  mx  + ,  by  division. 

I      That  is,  the  numerical  coefficient  of  x  is  the  same  as  the 
exponent  of  the  binomial  when  the  exponent  is  negative. 


BINOMIAL    THEOREM.  347 

Therefore,  whether  n  is  positive  or  negative,  integral 
or  fractional,  in  the  expansion  of  (l  +  a-)"  the  numerical 
coefficient  of  x  will  be  n. 

To  obtain  the  coefficients  of  x^,  x^,  etc.,  in  the  expansion 
of  (1 -\- xy\  we  separate  x  into  two  parts,  ?/  and  ^,  and 
write  the  expression  1  +  ic  in  two  ways  :  first,  retaining  the 
parts  of  x  in  connection  ;  and  secondly,  separating  them. 

We  then  compare  the  two  results. 

First  form : 

(i+x)"-(i+y+^r 

=  1  +  7.//+%^+ 6y+ 

+ + 

Second  form : 

(l+x)»  =  (l+y  +  ..)"  =  [(l  +  .r)  +  y]" 


,.)"  (^ 


(1 +  ,.)"!  + 


In  this  last  equality  it  is  assumed  that  the  power  of  a 
product  is  the  product  of  the  powers  of  the  factors,  what- 
ever be  the  exponent. 


■)■('+ fi-)' 


(1+. 


+  nz  +  n  (n  - 1)  zy  +  B  (u  -2)zif+ 

+ + 


348  ALGEBRA. 

Since  these  two  expansions  are  tlie  expansions  of  the 
same  thing,  in  the  same  form,  the  coefficients  of  z,  zy,  zi/, 
etc.,  must  be  the  same  in  both. 

In  the  first  expansion :  '          In  the  second  expansion : 

The  coefficient  of  z     is  n.  The  coefficient  of  z     is  n. 

The  coefficient  of  Z7j  is  2  B.  The  coefficient  of  zy  is  n  (?i  —  1). 

The  coefficient  of  zy^  is  3  C.  The  coefficient  of  zy^  is  B{n  — 2). 

Therefore,  ?i  =  ?i;  2B  =  n(7i  —  l)',  SC  =  B(n  —  2). 

Hence  ^=<!^ll).   ^^^(^-2)  ^n(^-l)(^-2). 
Hence,  i^        1X2  '   ^  3  1X2X3     '  '^^''• 

Accordingly,  whether  n  is  positive  or  negative,  integral 
or  fractional, 

431.  In  the  expansion  of  (l-\-x)^,  if  ti  is  a  positive 
integer,  the  numerator  of  the  last  factor  of  the  coefficient 
of  the  (r-]-l)th  term,  n  —  r  +  1,  will  be  equal  to  0  when 
7'  =  7i-\-l;  this  term,  therefore,  and  all  following  terms 
(for  they  will  also  have  this  factor)  will  vanish.  Hence, 
the  series  will  end  with  rth,  or  (n  -\-  l)th,  term.  But  if  n  is 
fractional  or  negative,  no  value  of  r  will  make  n — ?'+l=0, 
and  the  series  will  be  infinite.  Hence,  the  sign  =^  in  these 
cases  will  mean  •'  is  equal  to  the  limit  of  the  series.^^ 

432.  The  sum  of  the  numerical  coefficients  of  the  expan- 
sion of  (1  +  ^)"  is  equal  to  2";  for  the  sum  is  equal  to 
(1  + 1)",  as  appears  by  writing  1  for  x  in  (1  -|-ic)". 

433.  The  sum  of  the  positive  coefficients  is  equal  to  the 
sum  of  the  negative  coefficients  in  the  expansion  of  (1  —  x)"; 
as  appears  by  writing  1  for  x  in  (1  —  cc)". 

Note.  The  proof  in  §  430  is  taken  from  the  article  on  Algebra  in 
Encyclopaedia  Britannica. 


BINOMIAL    THEOREM.  349 


Expand  to  four  terms  : 

(1)  (i+^)i. 


1X2  1X2X3 

1+|X-|x2+/tX3 


^0^  —  2  ax 

2x)-i 


(2)  -7 
la^  —  2  ax 

=  Ml  I  ^  ,-H-i-i)/2xy _ -H-l-i)(-^-2)/2xy>j 

ai|       2a  1x2        \a/  1X2x3  \a/j 

ai(        2a      8a2      16a3^         j 
A  root  may  often  be  extracted  by  means  of  an  expansion. 

(3)  Extract  the  cube  root  of  344  to  six  decimal  places. 

344  =  343(14-,h)  =  73(14-jh). 

.•.^34i=7(l  +  jh)i, 

=  7(i  +  ^x,i,+^f^\jh)^+ y 

=  7  (1  +  0.000971815  -  0.000000944), 
=  7.00C796. 

(4)  Extract  the  fifth  root  of  3128  to  six  decimal  places. 

3128  =  55  + 3=  55(1  4- |y 
...</3T28  =  5(l  +  |y, 

=  (1  +  0.000192-0.000000073728  + ), 

=  6.000959. 


350  ALGEBRA. 


Exercise  121. 

Expand 

to  four  terms : 

1. 

(1  +  xy. 

(a  +  x)l 
(l-xr\ 

7. 
8. 

Q 

(2x-3y)-l. 

2. 
3. 

Vl-5x. 
1 

10. 

^(4.a'-Saxy 

4 

<}    ' 

^(1-32// 

5. 

(a'-xy. 

11. 

(i+x+xy. 

6. 

(x'  +  xy)-l 

12. 

(l~x  +  x^)i. 

13.  Find  the  rth  term  of  (a-{-x)K 

14.  Find  the  rth  term  of  (a  —  x)-^. 

15.  Find  V65  to  five  decimal  places. 

3, 

16.  Find  Vl^V  *^  ^"^^  decimal  places. 

17.  Find  Vl29  to  six  decimal  places. 

18.  Expand  (l  —  2x-\-3x^)-i  to  four  terms. 


19.    Find  the  coefficient  of  x"^  in  the  expansion  of        ,         3 


a-{'2xy 

(l  +  3.Ty 

20.    By  means  of  the  expansion  of  (1-f-^)^  show  that  the 
limit  of  the  series 

1  1       ■       1X3 1X3X5  .     /^ 

"^2      2X22"^2X3X2«      2X3X4X2^"^ ^        ' 


CHAPTER  XXIII. 
Chance. 

434.  If  an  event  can  happen  in  a  ways  and  fail  in  h 
ways,  and  if  all  these  ways  are  equally  likely  to  happen ; 
if,  also,  only  one  can  happen,  and  one  must  happen,  then 
the  mathematical  probability  or  chance  of  the  event  hap- 
pening is  expressed  by  the  fraction 

a 
a-]-h 

I.  The  probability  of  an  event  hapiyening  is  expressed  by 
the  fraction  whose  numerator  is  the  number  of  favorable 
ways,  and  denominator  the  whole  number  of  icays. 

Thus,  if  1  ball  is  drawn  from  a  bag  containing  3  white  balls  and 
9  black  balls,  the  chance  of  drawing  a  white  ball  is  y^^ ;  or,  as  it  is 
expressed,  one  chance  in  four. 

II.  The  probability  of  an  event  not  happening  is  expressed 
by  the  fraction  whose  numerator  is  the  number  of  unfavor- 
able ivays,  and  denominator  the  whole  number  of  ways. 

Thus  if  a  denote  the  number  of  favorable  ways,  and  h  the  number 
of  unfavorable  ways,  then  the  fraction  will  express  the  proba- 

bility of  the  event  not  happening.  If,  for  example,  1  ball  is  drawn 
from  a  bag  containing  3  white  and  9  black  balls,  the  chance  that  it 
will  not  be  a  white  ball  is  r\. 

-«.-     o,.  a       ,       b  . 

435.  Since  — r-r  H rT  =  lj 

a-\-b      a-\-o 

it  is  evident  that  the  chance  of  an  event  happening,  added 
to  the  chance  of  its  not  happening,  is  equal  to  1 ;    and, 


352  ALGEBRA. 

since  an  event  is  certain  to  happen  or  not  happen,  it  follows 
that  in  the  theory  of  chances 

III.  Certainty  is  expressed  by  unity. 

436.  Since  — --;  =  1 —7,  it  follows  that, 

a-\-b  a-\-o 

IV.  The  chance  of  an  event  not  happening  is  found  by 
subtracting  from  unity  the  chance  that  it  does  happjen. 

437.  If  the  number  of  favorable  ways  is  equal  to  the 
number  of  unfavorable  ways,  then 

a  a 


The  chance  of  the  event  =  — —7  =  ^7-  =  4- 

a  -\-b       2a       ^ 


This  is  expressed  by  saying  ''  the  event  is  as  likely  to 
happen  as  not,"  or  "  there  is  an  even  chance  for  the  event," 
or  "  the  odds  are  even  "  for  and  against  the  event. 

Thus,  in  tossing  a  cent,  there  is  an  even  chance  that  it  will  fall 
with  the  head  up. 

438.  If  a'>b,  the  chance  of  the  event  happening  is > ^. 
This  is  expressed  by  saying  ''  the  event  is  probable,"  or 

"the  odds  are  as  a  to  ^  in  favor  of  the  event." 

If  a<Cb,  the  chance  of  the  event  happening  is  <  ^. 
This  is  expressed  by  saying  "  the  event  is  improbable," 

or  "the  odds  are  as  ^  to  a  against  the  event." 

Thus,  the  odds  are  as  5  to  3  in  favor  of  drawing  a  white  ball  at 
the  first  trial  from  a  bag  containing  5  white  and  3  black  balls. 

Again,  since  a  die  has  6  faces,  on  one  of  which  is  an  ace,  the  cliance 
for  an  ace  the  first  throw  is  I ;  and  the  odds  are  5  to  1  against  an  ace. 

439.  It  may  be  shown  that, 

V.  If  there  are  several  events  of  which  only  one  can  hap- 
pen, the  chdnce  that  some  of  them  will  happen  is  the  sum 
of  their  respective  chances  of  happening. 


CHANCE.  353 

For,  let  a,  6,  c denote  the  number  of  ways  favorable  to  the  first, 

second,  third event  respectively ;  and  let  p  denote  the  whole  num- 
ber of  ways,  all  equally  probable,  and  of  which  one,  and  onZy  one,  must 
happen.     Then  the  chances  of  the  first,  second,  third, event  are 

-,   -,   -, respectively. 

P    P    P 

Since  there  are  a  +  6  +  c  + ways  favorable  to  some  one  or 

other  of  the  events  happening,  the  chance  in  favor  of  some  one  or 
other  of  the  events  is 

a+b+  c+ a  ,  b   ,  c   , 

or    -H 1 h  

P  P      P      P 

If,  for  example,  a  bag  contains  3  white,  4  black,  5  red,  and  6  green 

balls,  the  chance  of  drawing  at  the  first  trial  a  white  or  a  black  ball  is 

Ts  "^  A  ~  IS  >  ^^^  chance  of  drawing  a  white  or  a  black  or  a  red  ball 

is  y\  +  j\  +  fV  ~  tI  ;  ^^^  chance  of  drawing  a  white  or  a  black  or  a 

red  or  a  green  ball  is  ^^  +  A  +  is  +  ts  —  tI  —  1 5  ^^^^^  i^'  certainty. 

(1)  When  two  dice  are  thrown,  what  is  the  chance  of 

throwing  double  aces  ? 

Each  die  may  fall  in  any  one  of  6  ways ;  therefore  both  dice 
in  C  X  6  =  36  ways  (§  403).  Of  these  ways  only  one  will  give 
double  aces.     Hence,  the  chance  of  double  aces  =  ^\.   Ans. 

(2)  What  is  the  chance  of  throwing  doublets  in  a  single 

throw  with  two  dice  ? 

The  dice  may  fall  in  36  ways.     Of  these,  6  will  be  doublets. 
Hence,  the  chance  of  throwing  doublets  =  /g  =  \.  Ans. 

(3)  What  is  the  chance  of  throwing  a  six  and  a  five  by  a 

single  throw  of  two  dice  ? 

The  dice  may  fall  in  36  ways.  Of  tliese  ways  the  first  die  may 
turn  up  a  six  and  the  second  a  five,  or  the  first  may  turn  up  a 
five  and  the  second  a  six.     Hence,  the  chance  is  g^  =  y^.   An^. 

(4)  With  two  dice,  what  is  the  chance  of  making  a  throw 

so  that  one  and  only  one  die  may  turn  up  a  five  ? 

In  6  of  the  36  possible  ways  one  die  will  turn  up  a  five,  and 
the  other  also  will  turn  up  a  five  in  6  ways.  Two  of  these  12 
ways  will  be  double  fives;  so  that  there  are  10  ways  in  which  one 
die,  and  oyily  one,  will  turn  up  a, five,  and  the  chance  is  if.   Ans. 


354  ALGEBRA. 

(5)  What  is  the  chance   of   making   a   throw   that   will 

aviount  to  five  ? 

Of  the  36  possible  ways,  1  and  4,  4  and  1,  2  and  3,  3  and  2 
amount  to  jive.     Hence,  the  chance  is  /^  =  \.    Ans. 

(6)  In  a  single  throw  with  two  dice,  if  the  player  may 

count  the  number  on  one  of  the  dice,  or  the  sum  of 
the  numbers  on  the  two  dice,  what  is  the  chance  of 
throwing  five? 
The  chance  is  ^§  +  ^  +  gV  (double  fives),  that  is,  if.    Ans. 

(7)  If  A's  chance  of  winning  a  prize  is  i,  and  B's  \,  what 

is  the  chance  that  neither  will  obtain  a  prize  ? 

The  chance  that  one  will  win  is  |  +  |  =  /j.  Hence,  the 
chance  that  neither  will  win  is  1  —  j'j  =  W-    Ans. 

(8)  If  4  cards  are  drawn  from  a  pack  of  52  cards,  what  is 

the  chance  that  there  will  be  one  of  each  suit  ^ 

Four  cards  can  be  selected  (§415)  from  the  pack  in 

52  X  51  X  50  X  49      „^^  ^^^ 

— ; :^ —  =  270,725  ways. 

1X2X3X4  '  ^ 

But  4  cards  can  be  selected  so  as  to  be  one  of  each  suit  in 

13*  =  28,561  ways.  §  404. 

Hence,  the  chance  is  2W7V3  —  to?  nearly. 

(9)  If  4  cards  are  drawn  from  a  pack,  what  is  the  chance 

that  they  will  be  the  4  aces  ? 

There  are  270,725  ways  of  drawing  four  cards,  and  only  one 
way  of  drawing  the  four  aces.  Hence  the  chance  is  2707255  ^^ 
1  chance  in  270,725. 

(10)  Three  balls  are  to  be  drawn  from  an  urn  containing  5 

black,  o  red,  and  2  white  balls.     What  is  the  chance 
of  drawing  one  red  and  2  black  balls  ? 

10  X  9  X  8 

Three  balls  can  be  selected  from  the  whole  10  in 

1x^X0 

=  120  ways.    Also,  2  black  balls  can  be  selected  from  the  5  black 
balls  in =  10  ways,  and  1  red  ball  from  the  3  red  balls  in  3 

1    X    ii 


CHANCE.  355 

ways.  Hence,  1  red  and  2  black  balls  can  be  drawn  in  3  X  10 
=  30  ways.  That  is,  there  are  120  different  ways  of  drawing 
3  balls,  and  30  of  these  ways  give  1  red  and  2  black  balls. 

The  chance,  then,  of  1  red  and  2  black  balls  is  ^^^  =  i.    Ans. 

(11)  If  2  tickets  are  drawn  from  a  package  of  30  tickets, 

marked  1,  2,  3, ,  what  is  the  chance  that  both 

will  be  marked  with  odd  numbers  ? 

30  X  29 
Two  tickets  can  be  drawn  from  30  tickets  in  — j —  ways ; 

and  2  odd  numbers  can  be  drawn  from  the  15  odd  numbers  in 
15  X  14                _,           ^^       ,           .    15  X  14        .        . 

■■  ^  Q    ways.     Hence,  the  chance  is  5777-7...  =  ^'y-  ^^^- 

1  X  ^  ox)  X  ^y 

(12)  In  a  bag  are  5  white  and  4  black  balls.     If  they  are 

drawn  out  one  by  one,  what  is  the  chance  that  the 
first  will  be  white,  the  second  black,  and  so  on, 
alternately  ? 

The  9  balls  can  be  arranged  in  [9  ways.  The  5  white  balls 
can  be  arranged  in  the  odd  places,  and  the  4  black  balls  in  the 
even  places,  in  [5  X  [4  ways.  Hence  the  chance  of  alternate 
order  is  ^  X  |J 

—— —  =  ^^^.   Ans. 

(13)  From  a  bag  containing  10  balls,  4  are  drawn  and  re- 

placed, then  6  are  drawn.  Find  the  chance  that 
the  4  first  drawn  are  among  the  6  last  drawn. 

The  second  drawing  could  be  made  in 

[10 
j^- 210  ways. 

But  the  drawing  can  be  made  so  as  to  include  the  4  first  drawn 

^  [6 

^=15  ways. 

since  the  only  choice  consists  in  selecting  2  balls  from  the  6  not 
previously  drawn.     Hence,  the  chance  is  ^Vo  —  tt-   ^^*- 


356  ALGEBRA. 

(14)  The  chance  of  an  event  is  f .     What  are  the  odds  in 

favor  of  the  event  ? 

4  to  3.  Ans. 

(15)  The  odds  against  an  event  are  3  to  1.     What  is  the 

chance  of  the  event  ? 

i.   Ans. 

(16)  The  odds  against  an  event  are  m  to  n.     What  is  the 

chance  of  the  event  ? 

Ans. 


m  +  n 


(17)  If  4  coppers  are  tossed,  what  is  the  chance  that  ex- 

actly 2  will  turn  up  heads  ? 

Since  each  coin  may  fall  in  two  ways,  the  4  coins  may  fall 

in  2*  =  16  ways.     The  2  coins  to  turn  up  heads  can  be  selected 

4x3 

from   the  4  coins  in =  6  ways.     Hence,  the  chance  is 

1  X  ^ 

T6  ~  1 5  ^^^  ^^^  0^^^  ^^^  5  to  3  against  it. 

(18)  A  has  3  tickets  in  a  lottery  where  there  are  3  prizes 

and  6  blanks.     Find   his   chance  of  winning  one 
prize,  two  prizes,  three  prizes,  respectively. 

Q  X  8  X  T 
Three  tickets  can  be  selected  from  9  tickets  in 

1  X  2  X  o 

=  84  ways.     A  prize  ticket  can  be  selected  from  the  3  prize 

tickets  in  3  ways,  and  2  blanks  can  be  selected  from  the  6 

6x5 
blanks  in =15  ways ;    therefore,  1  prize  and  2  blank 

1  X   ii 

tickets  can  be  selected  in  3  X  15  =  45  ways.    Hence,  the  chance 

of  drawing  one  prize  is  ||. 

Again,  1  blank  and  2  prize  tickets  can  be  selected  in  6  X 

3x2 

=  18  ways.     Hence  the  chance  of  two  prizes  is  if. 

X  X  ^ 

Also,  the  3  prize  tickets  can  be  selected  in  only  1  way. 
Hence,  the  chance  of  drawing  three  prizes  is  ^^. 


CHANCE.  357 

(19)  What  is  the  chance  that  A  in  Ex.  (18)  wins  at  least 

one  prize? 

The  chance  is  ||  +  if  +  g^i  =  f  |  =  If.  For,  he  will  have  at 
least  one  prize  in  any  one  of  the  three  cases  given  in  (18). 

Or,  the  chance  may  be  found  in  this  way :  A  gets  a  prize 

unless  his  three  tickets  all  turn  out  blanks.     Three  tickets  can 

be  selected  from  the  whole  number  in  84  ways,  and  from  the  6 

0x5x4 
blanks  in  =  20  ways.     Hence,  the  chance  that  they 

1    X    Z    X    »5 

will  all  be  blanks  is  f|  =  2t  ;  and  the  chance  against  this  result 

is  1  -  A  =  2T- 

440.  If  a  person  is  to  receive  a  prize  in  case  a  particular 
event  liappens,  the  sum  of  money  for  which  he  may  equit- 
ably sell  his  chance  for  the  prize  is  called  his  expectation 
from  the  event. 

VI.  The  expectation  from  an  uncertain  event  is  the  product 
of  the  chance  that  the  event  will  happen  by  the  sum  to  he 

realized  in  case  the  event  happens. 

Thus,  if  there  is  a  lottery  with  40  tickets,  and  1  prize  worth  .^100, 
a  person  might  equitably  pay  $100  for  the  whole  40  tickets,  since  one 
of  them  is  sure  to  draw  the  )$100.  Now,  all  of  the  tickets  are  of  equal 
value  before  the  drawing ;  hence  the  value  of  each  ticket  is  -^^  of  i^lOO. 
The  value  of  5  tickets  is  f^  of  $100,  or  .^12.50  ;  that  is,  the  product  of 
the  chance  which  the  holder  of  5  tickets  has  of  winning  the  prize  and 
the  value  of  the  prize  ;  that  is,  ^^  of  $100. 

(20)  If  a  lottery  has  1  prize  of  ^50,  2  prizes  of  $5  each, 

4  prizes  of  f  1  each,  and  13  blanks,  what  is  the 
expectation  of  the  holder  of  1  ticket  ? 

The  chance  of  drawing  the  prize  of  |60  is  ^V?  and  the  expecta- 
tion is  ^^  of  $50  —  $2.50.  The  chance  of  drawing  a  prize  of 
$5  is  ^2^,  and  the  expectation  is  .f^  of  $5  =  $0.50.  The  chance 
of  drawing  a  prize  of  $1  is  2%,  and  the  expectation  is  2"^  of  $1 
=  $0.20.  Hence,  the  whole  expectation  is  $2.50  +  $0.50  + 
$0.20  =  $3.20.    Ans. 


358  ALGEBHA. 


Exercise  122. 

1.  If  I  throw  a  single  die,  what  is  the  chance  that  it  will 

turn  up : 
(i.)  An  ace  ? 
(ii.)  An  ace  or  a  two  ? 
(iii.)  Neither  an  ace  nor  a  two  ? 

2.  The  chance  of  a  plan  succeeding  is  i.     What  is  the 

chance  that  it  fails  ? 

3.  If  the  odds  are  10  to  1  against  an  event,  what  is  the 

probability  of  its  happening  ? 

4.  If  the  odds  are  5  to  2  in  favor  of  the  success  of  an 

experiment,    what   are   the    respective    chances    of 
success  or  failure  ? 

5.  The  chance  of  an  event  is  f.     Find  the  odds  for  or 

against  the  event. 

6.  What  is  the  chance  of  a  year,  not  a  leap-year,  having 

53  Sundays  ? 

7.  Two  numbers  are  chosen  at  random.     Find  the  chance 

that  their  sum  is  even. 

8.  If  4  cards  are  drawn  from  a  pack,  what  is  the  chance 

that  they  will  all  be  hearts  ? 

9.  If  10  persons  stand  in  a  line,  what  is  the  cliance  that 

2  assigned  persons  will  stand  together  ? 

10.  If  10  persons  form  a  ring,  what  is  the  chance  that  2 

assigned  persons  will  stand  together  ? 

11.  Show  that,  if  n  persons  sit  down  at  a  round  table,  the 

odds    against  2  particular   persons   sitting  next  to 
each  other  are  n  —  3  to  2. 


CHANCE. 


359 


12.  If  2  letters  are  selected  at  random  out  of  the  alphabet, 

what  is  the  chance  that  both  will  be  vowels  ? 

13.  Five  men,  A,  B,  C,  D,  E,  speak  at  a  meeting,  and  it  is 

known  that  A  speaks  before  B.  What  is  the  chance 
that  A  speaks  immediately  before  B  ? 

14.  A,  B,  C  have  equal  claims  for  a  prize.     A  says  to  B, 

"  You  and  I  will  draw  lots,  and  the  winner  shall 
draw  lots  with  C  for  the  prize."     Is  this  fair  ? 

15.  A  person  is   allowed  to  draw  2  tickets  from  a  bag 

containing  40  blank  tickets,  and  10  tickets  each 
entitling  the  holder  to  a  prize  of  $100.  What  is 
his  expectation  ? 

16.  One  of  two  events  must  happen.     If  the  chance  of  one 

is  I  of  that  of  the  other,  find  the  odds  on  the  first. 

17.  There  are  3  events,  A,  B,  C,  one  of  which  must  happen. 

The  odds  are  3  to  8  on  A,  and  2  to  5  on  B.  Find 
the  odds  on  C. 

18.  In  a  bag  are  7  white  and  5  red  balls.    Find  the  chance 

that  if  one  is  drawn  it  will  be  (i.)  white  or  (ii.)  red ; 
or,  if  two  are  drawn,  that  they  will  be  (i.)  both  white, 
(ii.)  both  red,  or  (iii.)  one  white  and  the  other  red. 

19.  If  3  cards  are  drawn  from  a  pack,  what  is  the  chance 

that  they  will  be  king,  queen,  and  knave  of  the 
same  suit? 

20.  A  general  orders  2  men  by  lot  out  of  100  mutineers  to 

be  shot ;  the  real  leaders  of  the  mutiny  being  10  in 
number.  Find  the  chance  (i.)  that  one  only,  (ii.) 
that  two,  of  the  leaders  will  be  shot. 

21.  Show  that  the  odds  are  8  to  1  against  throwing  9  in  a 

single  throw  with  2  dice. 


360  ALGEBRA. 

22.  Show  that  in  a  throw  with  3  dice  the  chance  of  either 

a  triplet  or  a  doublet  is  |. 

23.  In  a  bag  are  5  white  and  4  black  balls.    If  drawn  out, 

one  by  one,  what  is  the  chance  that  the  first  will  be 
white,  the  "second  black,  and  so  on,  alternately  ? 

24.  A  bag  contains  2  white  balls,  3  black  balls,  and  5  red 

balls.     If  4  balls  are  drawn,  find  the  chance  that 
there  shall  be  among  them  : 
(i.)  Both  the  white  balls, 
(ii.)  Two  only  of  the  black  balls, 
(iii.)  Two  at  least  of  the  red  balls. 

441.  A  series  of  events,  such  that  07ilij  one  of  them  can 
happen,  may  be  called  a  series  of  exclusive,  or  dependent, 
events. 

Two  or  more  events,  such  that  both  or  all  may  happen, 
are  called  non-exclusive,  or  independent,  events. 

Thus,  if  a  copper  be  thrown  twice  in  succession  it  may  fall  head 
up  both  times ;  and,  if  it  be  thrown  ten  times,  it  is  possible  for  it  to 
fall  head  up  each  time. 

442.  If  there  are  two  or  more  independent  events,  the 
occurrence  of  all  of  them  simultaneously  or  in  succession 
may  be  regarded  as  a  single  compound  event. 

Thus,  in  tossing  a  copper  twice,  the  event  of  its  falling  with  head 
up  at  both  trials  may  be  regarded  as  an  event  compounded  of  two 
simple  events ;  namely,  with  head  up  at  the  first  trial,  and  with  head 
up  at  the  second  trial. 

v/(l)    In  tossing  a  copper  twice,  what  is  the  chance  of  its 
falling  head  up  both  times  ? 

The  chance  of  a  head  at  each  trial  is  \.  If  these  separate 
chances  were  added  (according  to  Rule  V.),  the  result  would  be 
1 ;  that  is,  certainty;  a  result  obviously  false.  Rule  V.  applies 
only  to  dependent  or  exclusive  events.  In  this  case,  however, 
the  events  are  independent,  or  non-exclusive. 


CHANCE.  361 

Now,  each  time  the  copper  is  thrown,  it  can  fall  in  two  ways. 
Hence,  the  double  fall  can  occur  in  2  X  2  =  4  ways:  §  403. 

1.  Both  times  a  head, 

2.  First  time  a  head,  second  time  a  tail. 

3.  First  time  a  tail,  second  time  a  head. 

4.  Both  times  a  tail. 

Only  one  of  these  four  ways  gives  heads  both  times.  Hence, 
the  chance  of  heads  both  times  is  ^  =  i  x  ^  ;  that  is,  the  product 
of  the  separate  chances  of  a  head  at  each  trial. 

In  general, 

VII.  The  chance  that  two  independent  events  hoth  happen 
is  the  product  of  their  separate  chances  of  happening. 

For  the  product  of  the  denominators  of  the  separate  chances  is  the 
whole  number  of  ways  in  which  the  compound  event  can  happen ; 
and  the  product  of  the  numerators  is  the  number  of  ways  favorable 
to  its  happening, 

(2)  A  bag  contains  3  balls,  two  of  which  are  white; 
another  contains  6  balls,  five  of  which  are  white. 
If  a  person  draws  1  ball  from  each  bag,  what  is 
the  chance  that  hoth  balls  drawn  will  be  white  ? 

The  first  ball  can  be  drawn  in  3  ways  and  the  second  in  0 
ways.  Hence,  both  can  be  drawn  in  3x0=18  ways.  Also, 
the  first  ball  can  be  a  white  ball  in  2  ways,  and  the  second  in  5 
ways.  Hence,  they  can  be  both  ivhite  in  2  X  5  =  10  ways.  The 
chance  of  both  white  therefore  is  j§  =  f  x  |  ;  that  is,  the  product 
of  the  separate  chances  of  a  white  ball  at  each  trial, 

443.  In  like  manner,  whatever  the  number  of  simple 
events  that  unite  to  produce  a  compound  event,  it  may  be 
shown  that : 

VIII.  The  chance  of  a  compound  event  is  the  product 
of  the  separate  chances  of  the  simple  events  that  unite  to 
produce  it. 

NoTK.  It  is  important  not  to  confound  exclusive  events  with  non- 
exclusive, and  not  to  apply  Rule  V.  to  problems  to  which  Rule  VII. 
applies. 


362  ALGEBRA. 

,/ (3)  The  chance  that  A  can  solve  a  given  problem  is  f,  and 
the  chance  that  B  can  solve  it  is  ^^.  If  both  try, 
what  are  the  chances  (i.)  that  both  solve  it;  (ii.) 
that  A  solves  it  and  B  fails ;  (iii.)  that  A  fails  and 
B  solves  it ;  (iv.)  that  both  fail  ? 

A's  chance  of  success  is  f ,  A's  chance  of  failure  is  ^. 
B's  chance  of  success  is  j\,  B's  chance  of  failure  is  ^^. 
Therefore,  the  chance  of    (i. )  is  f  X  ^\  =  H  ; 

the  chance  of  (ii.)  is  f  X  -j?.  r=  ^4  . 

the  chance  of  (iii.)  is  -^  X  y\  =  -f-^  ; 

the  chance  of  (iv.)  is  ^  X  -^-^  =  ■^-^. 
The  sum  of  these  four  chances  is  |^f  +  if  +  /e  "I"  s'g  ~  1?  ^^  i^ 
ought  to  be,  since  1  of  the  4  results  is  certain  to  happen. 

/  (4)    In  Ex.  (3)  what  is  the  chance  that  the  problem  will  be 
solved  ? 

The  chance  that  both  fail  is  ^\.  Hence,  the  chance  that  both 
do  not  fail,  or  that  the  problem  will  be  solved,  is  1  —  /^  =  |f . 

(5)  There  are  3  bags,  the  first  containing  1  white,  and  1 
black  ball ;  the  second,  1  red  and  2  white  balls  ;  the 
third,  3  white  and  2  green  balls.  If  a  person  draw 
a  ball  from  each  bag,  what  is  the  chance  that  all 
three  balls  drawn  will  be  white  ? 
I  X  f  X  I  =  1.   Ans. 

*'    (6)    Under  the  conditions  of  the  last  problem,  what  is  the 
chance  that  no  one  of  the  balls  drawn  will  be  white  ? 
The  chances  of  failing  to  draw  a  white  ball  at  the  three  trials 
are  i,  i,  |,  respectively.     Therefore,  the  chance  of  failing  alto- 
gether is  ^  X  ^  X  I  =  ^J^,    Ans. 

(7)    What  is  the  chance  in  Ex.  (5)  of  drawing  at  least  one 
white  ball? 

One  white  ball  will  be  drawn  unless  all  three  trials  fail.  The 
chance  that  all  three  fail  is  ^^.  Therefore,  the  chance  of  draw- 
ing at  least  one  white  ball  is  1  —  ^-^  =  i|.   Ans. 


CHANCE.  363 

^8)    What  is  the  chance  in  Ex.  (5)  that  one,  and  only  one, 
white  ball  should  be  drawn  in  the  three  trials  ? 

The  chance  of  a  white  ball  from  the  first  bag  and  not 
from  the  others  is |  x  ^  x  ?  =  32^ 

The  chance  of  a  white  ball  from  the  second  bag  and 
not  from  the  others  is i  ><  i-  ><  1  =  i^ 

The  chance  of  a  white  ball  from  the  third  bag  and  not 
from  the  others  is i  >*^  i  >^  5  —  3% 

Therefore,  the  sum  of  these  chances  is ^^ 

</  (9)    When  6  coins  are  tossed,  what  is  the  chance  that  at 
least  one  will  fall  with  the  head  up  ? 
The  chance  that  all  will  fall  heads  down  is^x^x^^x^x^ 
X  ^  =  ^'i-      Hence   the  chance   that  this  will   not  happen  is 

,^10)    When  6  coins  are  tossed,  what  is  the  chance  that  one, 
and  only  one,  will  fall  with  the  head  up  ? 

The  chance  that  the  first  alone  falls  with  heads  up  is  |  X  | 
X|X^x^X^=j5'^;  the  chance  that  the  second  alone  falls 
with  the  head  up  is  -^^,  and  so  on. 

Hence,  the  chance  that  some  one,  and  only  one,  falls  head 
^V^-zh'^^\+^\  +  z\  +  ii-^h  =  z\  =  ^\-    ^ns. 

(11)  When  4  dice  are  thrown,  what  is  the  chance  that  two, 

and  only  tivo,  turn  up  aces  ? 

The  chance  that  any  particular  two  of  the  4  dice  turn  up 
aces,  and  the  other  two  something  else,  is  ^  X  i  X  a  X  §  =  T||^;. 

Now  the  number  of  ways  in  which  this  can  happen  is  the 
number  of  ways  in  which  two  dice  can  be  selected  from  4  dice, 
or  6  ways.  Hence,  the  chance  that  two,  and  only  two,  turn  up 
aces  is  6  X  y2  5^  =::  ^2^5^^    ^^5^ 

(12)  When  4  dice  are  thrown,  what  is  the  chance  that 

they  will  all  turn  up  alike  ? 

The  chance  that  the  first  and  second  turn  up  alike  is  ^. 
The  chance  that  the  third  turns  up  like  the  first  and  second 
isi. 

The  chance  that  the  fourth  turns  up  like  the  others  is  |. 
Hence,  the  chance  that  the.  four  turn  up  alike  is  j|^.   Ans, 


364  ALGEBRA. 

(13)  When  4  dice  are  thrown,  what  is  the  chance  that  two, 

and  only  two  of  them,  should  turn  up  alike  ? 

The  chajice  that  any  two  should  he  alike  is  ^,  the  chance 
that  the  third  should  be  different  is  |,  and  the  chance  that  the 
fourth  should  be  different  from  all  the  rest  is  |.  Hence,  the 
chance  that  a  pair  should  agree  while  the  others  should  differ 
from  the  pair  and  from  each  other  is  i  X  |  X  a  =  ^\^^-     The 

4x8 
pair  to  agree  may  be  selected  in  j——:  =  6  ways.     Hence,  the 

i  X  ^ 

total  chance  is  6  X  ^tg  —  f-    ^^^' 

(14)  When  4  dice  are  thrown,  what  is  the  chance  that 

they  should  all  fall  different  ? 


that  the  third  should  differ  from  both  the  first  and  the  second 
is  I,  and  that  the  fourth  should  differ  from  all  the  others  is  |. 
Hence,  the  required  chance  is  f  X  |  X  |  =  j^^.  Ans. 

(15)  A  single   die   is  thrown  until  it  turns  up   an  ace. 

What  is  the  chance  that  it  must  be  thrown  at  least 
10  times?  What  is  the  chance  that  it  must  be 
thrown  exact  If/  10  times  ? 

The  chance  of  failing  the  first  9  times  is  (Rules  IV.  and  VII.) 
(1)^.  This,  then,  is  the  chance  that  at  least  10  trials  must  be 
made.  Since  (1)^  is  the  chance  of  failing  the  first  9  trials,  and 
I  the  chance  of  success  the  next  trial ;  therefore  (Rule  VII)  (f )^ 
X  ^  is  the  chance  that  exactly  10  throws  must  be  made. 

(16)  What  is  the  chance  that  a  person  with  2  dice  will 

throw  double  aces  exactly  3  times  in  5  trials  ? 

The  chance  of  throwing  double  aces  at  any  particular  trial  is 
i  ^  ^  —  aV?  ^^^  of  failing  is  |f .  Hence,  the  chance  of  succeeding 
at  3  assigned  trials,  and  failing  at  the  other  2  trials,  is  {-^\)^  X  (f  f  )2. 
Now  double  aces  will  be  thrown  exactly  three  times  if  thrown  in 
any  set  of  3  trials  that  may  be  assigned  out  of  the  5  trials,  and 
fail  in  the  other  2  trials.  3  trials  can  be  assigned  out  of  5  trials 
in  ^^  f  ^  f  =  10  ways.     Hence,  the  chance  is  {j\y  X  (f  f  )2  x  10. 

1  X  2  X  o 


CHANCE.  365 

(17)  A  and  B  throw  with  a  single  die  alternately,  A  throw- 
ing first ;  and  the  one  who  throws  an  ace  first  is  to 
receive  a  prize  of  $10.  What  are  their  respective 
expectations  ? 

The  chance  for  the  prize  at  the  first  throw  is  ^  ;  at  the  second, 
f  X  ^ ;  at  the  third,  (|)2  x  i ;  at  the  fourth,  (g)^  x  ^  ;  and  so  on. 

As  A  has  the  first,  third,  etc.,  and  B  the  second,  fourth,  etc., 
throws, 

A's  chance 


~8o£J  +  (;5)»ofJ+ -3.  BO  that, 


B's  chance 

A's  expectation  is  y\  of  .$10  =  $5  j*V,  and 

B's  expectation  is  y\  of  -$10  =  '$4j\. 

(18)  A  and  B  play  at  a  game  that  cannot  be  a  drawn  game, 

and  on  an  average  A  wins  3  games  out  of  5  games. 
Out  of  5  games  what  is  the  chance  that  A  wins  at 
least  three  ? 

The  chance  that  A  wins  3  assigned  games  out  of  5  games  is 
(1)8  X  {i)'^  =  -siirr  The  3  games  may  be  assigned  in  10  ways. 
Hence,  A's  chance  for  3  games  is  10  x  ^'y^/j  =  ^xls- 

The  chance  that  A  wins  4  games,  and  B  the  other  game,  is 
■STin  X  6  =  -ij^s-      The  chance  that  A  wins  all  the  games  is 

For  A  to  win  at  least  3  games,  he  must  win  3,  4,  or  5  games. 
Hence,  A's  chance  for  at  least  3  games  is  ^J|§  +  /yV^  +  ^VA 

=   2133 

(19)  A's  skill  at  a  game,  which  cannot  be  a  drawn  game, 

is  to  B's  skill  as  3  to  4.    If  they  play  3  games,  what 
■  is  the  chance  that  A  will  win  more  games  than  B  ? 

Their  respective  chances  of  winning  a  particular  game  are 
f  and  f .  For  A  to  win  more  games  than  B,  he  must  win  all  3 
games  or  2  games.  The  chance  that  A  wins  all  three  is  (f)^  = 
■^j.  The  chance  that  A  wins  any  assigned  set  of  2  games  out 
of  the  3  games,  and  that  B  wins  the  other,  is  (^)'-^  X  |.  As  there 
are  3  ways  of  assigning  a  set  of  2  games  out  of  3,  the  chance 
that  A  wins  2  games,  and  B  the  other,  is  (f )2  x  |  X  3  =  ^ff . 
Hence,  the  chance  that  A  wins  more  than  B  is  ^£^  +  ^f  f  =  ||f . 


366  ALGEBRA. 

(20)  In  the  last  example,  find  B's  chance  of  winning  more 

games  than  A. 

B's  chance  of  winning  all  three  games  is  (f )3  =  /j^.  His 
chance  of  winning  2  games,  and  A  the  other  game,  is  (f  )2  x  f 
X  3  =  iff.     Hence,  his  chance  of  winning  more  games  than  A 

io     64     4-144—208 

Notice  that  A's  chance  added  to  B's  chance,  iff  +  f  Jf  =  1. 
Why  should  this  be  so  ? 

(21)  A  plays  a  set  of  games  (drawn  games  excluded)  with 

B,  his  chance  of  winning  a  single  game  being  to 
B's  as  3  :  2.     What  is  the  probability  : 
(i.)  That  A  will  win  4  games  at  least  out  of  7  ? 
(ii.)  That  he  will  win  4  games  before  B  wins  3  ? 

(1.)  A's  chance  of  winning  a  single  game  is  ?,  and  B's  chance 
is  |.  The  chance  that  A  wins  at  least  4  games  out  of  7  is  the 
sum  of  the  chances  that  he  wins  4,  5,  6,  or  7  games  out  of  7, 
in  any  possible  order. 

The  chance  that  he  wins  all  7  games  =  (?)'^. 

The  chance  that  he  wins  6  games       =  7  (§)^  X  |. 

The  chance  that  he  wins  5  games       =  21  (5)5  x  (|)2. 

The  chance  that  he  wins  4  games       =  35  (§)'*  X  (§)3. 

The  sum  of  these  values  =      ,'     .    Ans. 

6^ 

(ii.)  Here  the  chance  required  is  that  A  shall  win  at  least  4 

games ;  that  is,  4,  5,  or  6  games  out  of  6. 

The  chance  that  he  wins  all  6  games  =  (?)''. 

The  chance  that  he  wins  5  games       =  6  X  (?)5  X  |. 

The  chance  that  he  wins  4  games       =  15  X  (^)*  X  (|)2. 

The  sum  of  these  values  =  ~^^.   ^ns. 

5^ 

(22)  In  a  certain  locality  it  is  found  that,  on  the  average 

for  10  years,  out  of  100  persons  40  years  old  at  -the 
beginning  of  the  decade,  20  die  ;  out  of  100  per- 
sons 50  years  old,  30  die ;  and  out  of  100  persons 
60  years  old,  40  die.  What  is  the  odds  against  a 
person  40  years  old  living  30  years  longer  ? 


CHANCE.  367 

The  chance  that  he  dies  between  40  and  50  is  i  ;  that  he  lives 
till  50,  and  dies  between  50  and  60,  is  |  X  y3_  =  _6^  .  that  he  lives 
till  00,  and  dies  between  60  and  70,  is  |  X  y^  X  j\  =  j\^--.  Hence, 
the  chance  that  he  dies  between  40  and  70  is  ^  +  ^5  +  j%%  = 
^^^.  Therefore,  the  odds  against  his  living  for  30  years  are  83 
to  42,  or  about  2  to  1. 

(23)  A  is  40  years  old  and  B  50  years  old.     What  is  the 
probability  that  at  least  one  of  them  will  be  alive 
10  years  hence  ? 
The  chance  that  A  dies  is  ^,  and  the  chance  that  B  dies  is  -^^j. 

Hence,  the  chance  that  both  die  is  1  X  ^%  =  /^ ;  and  the  chance 

that  one  at  least  will  be  alive  is  1  —  -^^  =  |^. 

444.  Cases  often  occur  where  the  simple  events  which 
unite  to  form  the  compound  event  are  so  related  that  the 
happening  of  one  of  them  alters  the  chances  of  the  others. 

i/(l)    What  is  the  chance  of  drawing  in  succession  2  vowels 

from  the  alphabet  ? 

The  chance  of  drawing  a  vowel  the  first  time  is  ^''^ ;  but,  if 
one  vowel  is  drawn,  the  chance  of  drawing  another  is  /j.  Hence, 
(Rule  VII.)  the  required  chance  is  ^%  X  A  —  sS- 

J^)  A  bag  contains  5  white  and  6  black  balls.  What  is 
the  chance  of  drawing  5  times  in  succession  a  white 
ball,  the  balls  drawn  not  being  replaced  ? 

fS)  What  would  have  been  tlie  chance  in  the  last  example 
if  after  each  drawing  the  ball  had  been  replaced  ? 

(4)  If  the  chance  of  an  event  at  first  is  as  a  to  b,  and  if 
whenever  it  happens,  the  number  of  favorable  ways, 
as  well  as  the  whole  number  of  ways,  is  diminished 
by  unity ;  find  the  chance  that  the  event  will  occur 
n  times  in  succession. 

a(a-l)(a-2) (a-n  +  1). 

6(6-l)(6-2) (6-n+l) 


368  ALGEBRA. 

y  (5)    A  bag  contains  5  white  and  6  black  balls.     If  5  balls 
are  drawn  in  succession,  and  no  one  of  them  replaced, 
what  is  the  probability  that  the  first  three  will  be 
white,  and  the  fourth  and  fifth  black  ? 
The  separate  chances  for  the  5  simple  events  are  respectively 
TT»  ro'  h  h  f  •     Hence,  the  chance  for  the  compound  event  is 
tVXtVxIX!  xf  =  Tf?-  ^ns. 
/(Q)    Find  the  probability  in  the  last  example  that  the  5 
balls  drawn  Avill  be  3  white  and  2  black  balls. 

Here  the  chance  required  is  that  3  wliite  and  2  black  should 
be  drawn  not  in  any  assigned  order,  as  in  the  last  case,  but  in 
any  possible  order.  Now  5  things,  of  which  3  are  alike  and  the 
other  2  alike,  may  be  arranged  (§  411)  in 

—^-10  ways. 


(7)  Find  the  respective  probabilities  in  Examples  (5)  and 

(6)  if  after  each  drawing  the  ball  is  replaced. 

In  Ex.  (5),     if'j)^  X  (j«^)2.    Ans. 

In  Ex.  (6),     10  X  (^\)3  X  (/j-)2.  Ans. 

(8)  A  purse  contains  9  silver  dollars  and  1  gold  eagle,  and 

another  contains  10  silver  dollars.  If  9  coins  are 
taken  out  of  the  first  purse  and  put  into  the  second, 
and  then  9  coins  are  taken  out  of  the  second  and 
put  into  the  first  purse,  which  purse  now  is  the 
more  likely  to  contain  the  gold  coin  ? 

The  gold  eagle  will  not  be  in  the  second  purse  unless  it  (i.) 
was  among  the  9  coins  taken  out  of  the  first  and  put  into  the 
second  purse,  (ii.)  and  not  among  the  9  coins  taken  out  of  the 
second  and  put  into  the  first  purse.  The  chance  of  (i.)  is  j%,  and 
when  (i.)  has  happened  the  chance  of  (ii.)  is  }f.  Hence,  the 
chance  of  both  happening  is  y^  X  i|  =  j\.  Therefore,  the  chance 
that  the  eagle  is  in  the  second  purse  is  j\,  and  the  chance  that 
it  is  in  the  first  purse  is  1  —  f'g  =  -{~°.  Since  [^  is  greater  than 
Y^^,  therefore  the  gold  coin  is  more  likely  to  be  in  the  first  purse. 


CHANCE.  369 

(9)  In  a  bag  are  2  red  and  3  white  balls.  A  is  to  draw  a 
ball,  then  B,  and  so  on  alternately ;  and  whichever 
draws  a  white  ball  first  is  to  receive  $10.  Find 
their  expectations. 

A's  chance  of  drawing  a  white  ball  at  the  first  trial  is  |.  B's 
chance  of  having  a  trial  is  equal  to  A's  chance  of  drawing  a  red 
ball  =  f .  In  case  A  drew  a  red  ball  there  would  be  1  red  and  3 
white  balls  left  in  the  bag,  and  B's  chance  of  drawing  a  white 
ball  would  be  f.  Hence,  B's  chance  of  having  the  trial  and 
drawing  a  white  ball  is  f  X  f  =  f'j ;  and  B's  chance  of  drawing 
a  red  ball  is  ?  x  i=  y'o- 

A's  chance  of  having  a  second  trial  is  equal  to  B's  chance  of 
drawing  a  red  ball  =  ■^^.  In  case  B  drew  a  red  ball  there  would 
be  3  white  balls  left,  and  A's  chance  of  drawing  a  white  ball 
would  be  certainty,  or  1. 

A's  chance,  therefore,  is  f  +  yV  ~  tV  5  ^.nd  B's  chance  is  f^. 

A's  expectation,  then,  is  $7,  and  B's  §3. 

445.  In  general,  when  it  is  required  to  find  wliich  of 
two  doubtful  events  is  more  likely  to  happen,  it  is  neces- 
sary to  find  their  respective  chances,  and  then  to  compare 
the  results  obtained. 

(1)  In  one  throw  with  two  dice  which  sum  is  more  likely 

to  be  thrown,  9  or  12  ? 

Out  of  the  30  possible  ways  of  falling,  four  give  the  sum  0 
(namely,  6  +  3,  3  +  0,  5  +  4,  4  +  5),  and  only  one  way  gives  12 
(namely,  G  +  0).  Hence,  the  chance  of  throwing  9  mfour  times 
as  good  as  that  of  throwing  12. 

(2)  With  three  dice    wliat   are   the    relative   chances   of 

throwing  a  doublet  and  a  triplet? 
The  chance  of  throwing  a  doublet  is 


3X 


6X1X5 


=  ¥i 


03 

The  chance  of  throwing  a  triplet  is 
6  X  1 X  1_   ^ 

Hence,  the  chance  of  a  doublet  is  15  times  that  of  a  triplet. 


370  ALGEBRA. 

(3)  A  bag  contains  1  black  and  4  white  balls,  and  another 

bag  contains  7  black  and  3  white  balls.  If  a  person 
draws  a  ball  from  one  of  the  bags,  (i.)  what  is  the 
chance  that  it  be  a  white  ball  ?  (ii.)  what  is  the 
ratio  of  the  chance  of  its  being  drawn  from  the  first 
bag  to  that  of  its  being  drawn  from  the  second  bag  ? 

The  person  (so  far  as  we  know)  is  as  likely  to  choose  one  bag 
as  the  other.  Hence,  the  chance  of  his  choosing  the  first  bag  is 
^ ;  and  the  cliance  of  his  drawing  a  wliite  ball  from  the  first  bag 
is  ^.  Therefore,  the  chance  of  drawing  a  white  ball  from  the 
first  bag  is  |  X  |  =  |.  In  the  same  way,  the  chance  of  drawing 
a  white  ball  from  the  second  bag  is  found  to  be  ^  X  /^  =  j'o- 

Therefore,  the  chance  of  drawing  a  white  ball  is  |  +  ^^  —  III 
and  the  ratio  of  the  separate  chances  is  8 :  3. 

(4)  Suppose  in  the  last  example  that  at  the  first  trial  a 

white  ball  is  actually  drawn.     What  are  now  the 

chances  that  it  came  from  the  first  bag,  and  from 

the  second,  respectively  ? 

Let  X  and  y  represent  the  chances  required. 

Then,  by  Ex.  (3),  -  =  7/ 

?/       3 

Also,  x-\-  y  =  1, 

since  the  ball  must  have  come  from  one  or  the  other  bag. 

The  solution  of  these  equations  gives 

x=  j^j,     and     y  =  j\. 

446.  From  Examples  (3)  and  (4)  it  will  be  seen  that, 
IX.  If  a  doubtful  event  may  happen  in  some  one  of  several 
ivays,  the  actual  Jtappening  of  the  event  changes  its  proba- 
bilitt/,  and  the  separate  2^ ^'ob abilities  of  the  several  ways  of 
hapj^eniny,  in  the  same  ratio  ;  and  this  ratio  is  the  reciprocal 
of  the  fraction  that  expresses  the  chance  of  the  event  before 
it  actually  happens. 

Thus,  in  Ex.  (3),  the  chance  of  the  event  hefore  it  happened,  and 
the  chance  of  the  two  separate  ways  of  happening,  were  found  to  be 


CHANCE.  371 

kh  /o'  io  y  ^^  ^^-  (4)  i*  w^  shown  that  after  the  event  happened  these 
chances  became  1,  j^^-,  /j,  respectively.  The  values  1,  j\,  y^j-  are  ob- 
tained by  multiplying  ii,  /q,  j'o-  by  f  J  ;  that  is,  by  the  reciprocal  of  ^i. 

Evidently,  the  happening  of  the  event  must  change  to  unity  the 
chance  of  the  event ;  and  must  therefore  increase  to  unity  the  sum  of 
the  separate  chances. 

So  long,  however,  as  the  only  additional  knowledge  about  the  event 
is  the  fact  that  it  has  happened,  the  relative  probabilities  of  the  sepa- 
rate ways  of  happening  remain  unchanged.  Therefore,  the  several 
fractions  which  before  expressed  the  probabilities  of  the  separate  ways 
of  happening  must  now  be  multiplied  by  the  same  factor,  and  that 
factor  is  the  reciprocal  of  the  fraction  that  expressed  the  probability 
of  the  event  before  it  happened. 

Exercise  123. 

1.  The  chance  that  A  can  solve  a  certain  problem  is  i,  and 

the  chance  that  B  can  solve  it  is  f.     What  is  the 
chance  that  the  problem  will  be  solved  if  both  try  ? 

2.  What  is  the  chance  of  throwing  at  least  one  ace  in  2 

throws  with  one  die  ? 

3.  If  w  coins  are  tossed  up,  what  is  the  chance  that  one, 

and  only  one,  will  turn  up  head  ? 

4.  What  is  the  chance  of  throwing  double  sixes  at  least 

once  in  3  throws  with  2  dice  ? 

5.  A  copper  is  tossed  3  times.     Find  the  odds  that  it  will 

fall: 

(i.)  One  head  and  two  tails  without  regard  to  order, 
(ii.)  Head,  tail,  head. 

6.  If  a  copper  is  tossed  4  times,  find  the  odds  that  it  will 

fall  2  heads  and  2  tails  sooner  than  4  heads. 

7.  If  from  a  lottery  of  30  tickets,  marked  1,  2,  3, ,  four 

tickets  are  drawn,  what  is  the  chance  that  1  and  2 
will  be  among  them  ? 


372  ALGEBRA. 

8.  If  2  coppers  are  tossed  3  times,  find  the  odds  that  they 

will  fall  2  heads  and  4  tails. 

9.  There  are  10  tickets,  five  of  which  are  numbered  1,  2, 

3,  4,  5,  and  the  other  five  are  blank.  Find  the 
chance  that  the  sum  of  the  numbers  on  the  tickets 
drawn  in  3  trials  will  be  10,  one  ticket  being  drawn 
and  then  replaced  at  each  trial  ? 

10.  Find  the  chance  in  Ex.  9  if  the  tickets  are  not  replaced. 

11.  A  bag  contains  4  white  and  6  red  balls.     A,  B  and  C 

draw  each  a  ball,  in  order,  replacing.  Find  the 
chance  that  they  have  drawn  : 

(i.)  Each  a  white  ball. 

(ii.)  A  and  B  white,  C  red. 
(iii.)  Two  white  and  one  red. 

12.  Find  the  answer  to  Ex.  11  if  the  balls  are  not  replaced. 

13.  A  draws  4  times  from  a  bag  containing  2  white  and  8 

black  balls,  replacing.  Find  the  chance  that  he 
has  drawn : 

(i.)  Two  white,  two  black. 

(ii.)  Not  less  than  two  white, 
(iii.)  Not  more  than  two  white, 
(iv.)  One  white,  three  black. 

14.  Find  the  odds  against  throwing  one  of  the  two  numbers 

7  or  11  in  a  single  throw  with  2  dice. 

15.  If  a  copper  is  tossed  5  times,  what  is  the  chance  that 

it  will  fall  heads  either  2  times  or  else  3  times  ? 

16.  Find  the  same  chance  if  the  copper  is  tossed  6  times. 

17.  In  one  bag  are  10  balls  and  in  another  6 ;  and  in  each 

bag  the  balls  are  marked  1,  2,  3,  etc.  What  is  the 
chance  that  on  drawing  one  ball  from  each  bag  the 
two  balls  will  have  the  same  number  ? 


CHANCE.  373 

18.  A  bag  contains  n  balls.     A  person  takes  out  one  ball 

and  then  replaces  it.  He  does  this  n  times.  What 
is  the  chance  that  he  has  had  in  his  hand  every  ball 
in  the  bag  ? 

19.  If  on  an  average  9  ships  out  of  10  return  safe  to  port, 

what  is  the  chance  that  out  of  5  ships  expected  at 
least  3  will  return  ? 

20.  What  is  the  chance  of  throwing  double  sixes  at  least 

once  in  3  throws  with  a  pair  of  dice  ? 

21.  What  is  the  chance  of  throwing  15  in  one  throw  with 

3  dice  ? 

22.  In  5  throws  with  a  single  die  what  is  the  chance  of 

throwing  an  ace : 

(i.)  Three  times  exactly? 

(ii.)  Not  less  than  three  times? 
(iii.)  Not  more  than  three  times  ? 

23.  In  a  bag  are  3  white,  5  red,  and  7  black  balls,  and  a 

person   draws   three   times,   replacing.      Find   the 
chance  that  he  has  drawn: 
(i.)  A  ball  of  each  color, 
(ii.)  Two  white,  one  red. 
(iii.)  Three  red. 
(iv.)  Two  red,  one  black. 

24.  A  and  B  play  at  chess,  and  A  wins  on  an  average  2 

games  out  of  3.  Find  the  chance  of  A's  winning 
exactly  4  games  out  of  the  first  6,  drawn  games 
being  disregarded. 

25.  A  and  B  engage  In  a  game  in  which  A's  skill  is  to  B's 

as  2:3.  Find  the  chance  of  A's  winning  at  least  2 
games  out  of  the  first  5,  drawn  games  not  being 
counted. 


374  ALGEBRA. 

26.  The  skill  of  A  is  double  that  of  B.     Find  the  odds 

against  A's  winning  4  games  before  B  wins  2. 

27.  If  B's  skill  in  a  certain  game  is  equal  to  three-fifths  of 

A's,  find  A's  chance  of  winning  5  games  out  of  8. 

28.  A  bag  contains  4  red  balls  and  2  others,  each  of  which 

is  equally  likely  to  be  red  or  white.  Three  times 
in  succession  a  ball  is  drawn  and  replaced.  Find 
the  chance  that  all  the  drawn  balls  are  red. 

29.  A  man  has  left  his  umbrella  in  one  of  3  shops  which 

he  visited  in  succession.  He  is  in  the  habit  of 
leaving  it,  on  an  average,  once  every  4  times  that 
he  goes  to  a  shop.  Find  the  chance  that  he  left  it 
in  the  first,  second,  and  third  shops,  respectively. 

30.  A  bets  B  $10  to  $1  that  he  will  throw  heads  at  least 

once  in  3  trials.  What  is  B's  expectation  ?  What 
would  have  been  a  fair  bet  ? 

31.  A  draws  5  times  (replacing)  from  a  bag  containing  3 

white  and  7  black  balls;  every  time  he  draws  a 
white  ball  he  is  to  receive  $1,  and  every  time  he 
draws  a  black  ball  he  is  to  pay  50  cents.  What  is 
his  expectation  ? 

32.  From  a  bag   containing   2    eagles,   3  dollars,   and   3 

quarter-dollars,  A  is  to  draw  one  coin  and  then  B 
three  coins ;  and  A,  B,  and  C  are  to  divide  equally 
the  value  of  the  remainder.  What  are  their  expec- 
tations ? 

33.  A,  B,  and  C,  staking  each  $5,  draw  from  a  bag  in  which 

are  4  white  and  6  black  balls,  each  drawing  in  order, 
and  the  whole  sum  is  to  be  received  by  him  who  first 
draws  a  white  ball.     What  are  their  expectations  ? 
(i.)  Replacing  the  balls, 
(ii.)  Not  replacing  the  balls. 


CHAPTEE    XXIV. 
Formulas. 

Simple  Interest. 

447.  If  in  Interest 

The  principal  is  represented  by  P, 

interest  on  $1  for  one  year  by  r, 
amount  of  $1  for  one  year  by  R, 
number  of  years  by  tj, 

amount  of  P  for  n  years  by         A. 
Then  ^  =  1  +  r, 

Simple  interest  on  P  for  a  year    =  Pr, 
Amount  of  P  for  a  year  =  Pi?, 

Simple  interest  on  P  for  n  years  =  Pnr, 
Amount  of  P  for  n  years  =  P(l  -|-  nr). 

That  is  ^  =  P(l  +  7ir). 

448.  When  any  three  of  the  quantities  A,  P,  n,  r  are 
given,  the  fourth  may  be  found. 

Ex.    Required  the  rate  when  $500  in  4  years  at  simple 
interest  amounts  to  $610. 

Here  r  is  required,  A,  P,  n  are  given. 
A  =  P{l  +  nr), 
or  A  =  P  +  Pnr. 

.'.  Pnr  =  A-  P. 

Pn  2000 

5^  per  cent.   Ans. 


376  ALGEBRA. 

449.  Since  P  will  in  n  years  amount  to  A,  it  is  evident 
that  P  at  the  present  time  may  be  considered  equivalent  in 
value  to  A  due  at  the  end  of  n  years ;  so  that  P  may  be 
regarded  as  the  present  worth  of  a  given  future  sum  A. 

Ex.    Find  the  present  worth  of  f  600,  due  in  2  years,  the 
rate  of  interest  being  6  per  cent. 

^  =  P  (1  4-  nr). 

1  +  nr      1  +  0.12 


Compound  Interest. 

450.    When  compound  interest  is  reckoned  payable  annu- 
ally, 

The  amount  of  P  dollars  in 

1  year  is       P{l-\-r)  =  PR, 

2  years  is  PR  (1  +  r)  =  PK\ 
n  years  =  PR^. 

That  is,  A  =  PR\ 

A 
Hence,  also,  P  =  — • 

When  compound  interest  is  payable  semi-annually^ 
The  amount  of  P  dollars  in 


iyear    =p(l  +  0, 
lyear    =p('l  +  0  , 

■^  "^  2/    ' 
That  is,  A  =  pI\+  ^\    • 

When  the  interest  is  payable  quarterly, 


FORMULAS.  377 

When  the  interest  is  payable  monthly, 

/         r  \i2» 

^  =  ^('  +  ^)  • 

When  interest  is  payable  q  times  a  year, 

Ex.    Find  the  present  worth  of  $500,  due  in  4  years,  at 
5  per  cent  compound  interest. 

Sinking  Funds. 

451.    If  the  sum  set  apart  at  the  end  of  each  year  to  be 
put  at  compound  interest  is  represented  by  S,  then, 
The  sum  at  the  end  of  the 

first  year      =  S, 
second  year  =  *SH-  SB, 
third  year    =S-\-SE+SB^ 

nth  year       =S-\-SE-\-  Sli-  + +  SIi^\ 

That  is,  the  amount  A  =  S-\-  SE-\-SE^-{- +  SE"-\ 

.-.  AR  =  SR-\-SR^-[-SE^+ +  SE". 

,\AE-A  =  SE'^-S. 


.'.A 


_S{E--1) 


or,  A 

r 

(1)    If  $10,000  be  set  apart  annually,  and  put  at  6  per  cent 
compound  interest  for  10  years,  what  will  be  the 

amount  ? 

^  S  (fi»  -  1)  _  $10,000  (l.OGi^  -  1) 
r  0.00 

By  logarithms  the  amount  is  found  to  be  •'§131,740  {Tiearly). 


378  ALGEBRA. 

(2)    A  county  owes  $60,000.     What  sum  must  be  set  apart 
annually,  as  a  sinking  fund,  to  cancel  the  debt  in  10 
years,  provided  money  is  worth  6  per  cent? 
„         Ar         .'$60,000  X  0.06       .  .__  .        ,  . 

Note.    The  amount  of  tax  required  yearly  is  $3600  for  the  interest 
and  .^4555  for  the  sinking  fund  ;  that  is,  $8155. 

Annuities. 

452.   A  sum  of  money  that  is  payable  yearly,  or  in  parts 
at  fixed  periods  in  the  year,  is  called  an  annuity. 

I.  To  find  the  amount  of  an  unpaid  annuity  when  the 
interest,  time,  and  rate  per  cent  are  given. 

The  sum  due  at  the  end  of  the 
first  year      =  S, 
second  year  =  /S^+  SB, 
third  year    =S+SR-{-SR^, 
nth  year        =S+SR^  SR^-\- +  SR^-\ 

That  is,  ^  =  ^(^^. 

r 

Ex.  An  annuity  of  $1200  was  unpaid  for  6  years.  What 
was  the  amount  due  if  interest  is  reckoned  at  6  per 
cent? 

_  S{Bn-\)  _  $1200(1.066-1)  _ 
^  -  —r         -  0:06  -  ^^"^^' 

II.  To  find  the  present  worth  of  a7i  annuity  when  the  time 
it  is  to  continue  and  the  rate  per  cent  are  given. 

Let  P  denote  the  present  worth.      Then  the  amount  of  P  for  n 
years  will  be  equal  to  A^  the  amount  of  the  annuity  for  n  years. 
But  the  amount  of  P  for  n  years 

=  P  (1  +  r)«  =  PE«, 

and  A  =  -^ — —-'■  §  451. 

K  —  i 

ji-l 


IP'       R  —  1 
This  equation  may  be  written 


FORMULAS.  379 


.■.P=#x«" 


A'  -  1  E« 

If  tlie  annuity  is  perpetual,  the  fraction 

approaches  to  unity  as  its  limit. 

.-.  P  =  hmit  of  ~ :  X  — -- —  =  — =  — 

li—  I         IP'         li—1      r 

(1)  Find  the  present  worth  of  an  annual  pension  of  $105, 

for  5  years,  at  4  per  cent  interest. 

^  ~  li"       U-1 

.•$105       1.045  —  1      ^^^^  ^ 

(2)  Find  the  present  worth  of  a  perpetual  scholarship  that 

pays  $300  annually,  at  6  per  cent  interest. 

-f=fi— ■ 

III.  To  fitul  the  present  worth  of  an  annuitij  that  begins 
in  a  given  number  of  years,  when  the  time  it  is  to  continue 
and  the  rate  per  cent  are  given. 

Let  p  denote  the  number  of  years  before  the  annuity  begins,  and 
q  tlie  number  of  years  the  annuity  is  to  continue. 

Tlien  the  present  worth  of  the  annuity  to  the  time  it  terminates  is 
S         Rp+1  -  1 
Rj'+i  ^     R-l     ' 
and  the  present  worth  of  the  annuity  to  the  time  it  begins  is 

Rp       R-l 
Hence, 


380  ALGEBRA. 

If  the  annuity  is  to  begin  at  the  end  of  p  years,  and  to  be  perpetual, 
the  formula 

BP+1      B  —  1 

S  Ri-1 

Ei  —  l 
And  since  the  limit  of  — - —  is  unity, 

P  =  the  limit  of  -—4 — 77  X  ^'^  ~  ^      '       ^ 


Rp{R-l)  Ri  Rp{R-l) 

(1)  Find  the  present  worth  of  an  annuity  of  $5000,  to 
begin  in  6  years,  and  to  continue  12  years,  at  6  per 
cent  interest. 


Rp+Q        R  —  1 
$5000       1.0612  -  1 
1.0618  ^       0.06 


=  i$29,550. 


(2)    Find  the    present  worth  of  a   perpetual  annuity  of 
$1000,  to  begin  in  3  years,  at  4  per  cent  interest. 

S  _       $1000       -^^ 

IV.    To  find  the  annuity  when  the  present  worthy  the  time, 
and  the  rate  per  cent  are  given. 

S{Rn-l) 
R^R  —  1) 

.         _  PRnjR-l)  _  Rn 

(1)    What  annuity  for  5  years  will  $4675  give  when  inter- 
est is  reckoned  at  4  per  cent  ? 

K«  1  045 

S=PrX  ^^^-^  =  $4675  X  0.04  X  ^^^^:^  =  $1050. 


FORMULAS.  381 

Life  Insurance. 

463.  In  order  that  a  certain  sum  may  be  secured,  to  be 
payable  at  the  death  of  a  person,  he  pays  yearly  a  fixed 
premium. 

If  P  denotes  the  premium  to  be  paid  for  n  years  to  insure  an 
amount  A,  to  be  paid  immediately  after  the  last  premium,  then 

^  =  IL(|— U.  5  451. 

.       ^^(f^-1)^     At    . 
R^-\         i2«  -  1 

If  A  is  to  be  paid  a  year  after  the  last  premium,  then 
A{R-\)  ^        Ar       _ 
R  {R"  -  1)       R  {R»  -  1) 

Note.  In  the  calculation  of  life  insurance  it  is  necessary  to 
employ  tables  which  show  for  any  age  the  probable  duration  of  life. 


Bonds. 

464.  If  P  denote  the  price  of  a  bond  that  has  n  years  to 
run,  and  bears  ?•  per  cent  interest,  S  the  face  of  the  bond, 
and  q  the  current  rate  of  interest,  what  interest  on  his 
investment  will  a  purchaser  of  such  a  bond  receive  ? 

Let  X  denote  the  rate  of  interest  on  the  investment. 

Then  P  (1  +  x)'»  is  the  value  of  the  purchase  money  at  the  end  of 
n  years. 

Sr  (1  +  q)«-i  +  Sr  {1  +  7)»-2  +  -\-  Sr -\-  S  is  the  amount  of  the 

money  received  on  the  bond  if  the  interest  received  from  the  bond  is 
put  immediately  at  compound  interest  at  q  per  cent. 

But  Sr{l  +  g)«-i  +  Sr  (1  +  q)^-^+ +  Sr+S=S+  ^^[^^+g)"~^3- 

Sq-\-Sr{l  +  q)'^-Sr\\ 
Pq  )' 


=c 


382 


ALGEBRA. 


(1)  What  interest  does  a  person  receive  on  his  investment 

if  he  buys  a  4  per  cent  bond,  at  114,  that  has  26 
years  to  run,  and  if  money  is  worth  3|-  per  cent  ? 

^  ^  "^  ~  V       3.99       ; 

By  logarithms,  l-\-x=  1.033. 

That  is,  the  purchaser  will  receive  3^  per  cent  for  his  money. 

(2)  At  what  price  must  7  per  cent  bonds  be  bought,  run- 

ning 12  years,  with  the  interest  payable  semi-annu- 
ally, in  order  that  the  purchaser  may  receive  on  his 
investment  5  per  cent,  interest  semi-annual,  which  is 
the  current  rate  of  interest  ? 

p  Sg  +  Sril  +  q)n-Sr^ 

q 
.  p^  Sq  +  Sr  (1  4-  qY  -  Sr 
q{l  +  x)« 
In  this  case  S  =  100  ;  and,  as  the  interest  is  semi-annual, 
q  =  0.025,  r  -  0.035,  n  =  24,  x  =  0.025. 
2.5+3.5(1.025)24-3.5 
Hence,  i^  -  0.025(1.025)'^* 

By  logarithms,     P  =  118. 

Exercise  124. 

1.  In  how  many  years  will  f  100  amount  to  $1050,  at  5 

per  cent  compound  interest  ? 

2.  In  how  many  years  will  %A  amount  to  f  i?  (i.)  at  sim- 

ple interest,  (ii.)  at  compound  interest,  r  and  R  being 
used  in  their  usual  sense  ? 

3.  Find  the  difference  (to  five  places  of  decimals)  between 

the  amount  of  $1  in  2  years,  at  6  per  cent  compound 
interest,  according  as  the  interest  is  due  yearly  or 
monthly. 

4.  At  5  per  cent,  find  the  amount  of  an  annuity  A  which 

has  been  unpaid  for  4  years. 


FORMULAS.  383 

5.  Find  the  present  value  of  an  annuity  of  ^100  for  5 

years,  reckoning  interest  at  4  per  cent. 

6.  A  perpetual  annuity  of  $1000  is  to  be  purchased,  to 

begin  at  the  end  of  10  years.  If  interest  is  reckoned 
at  3^  per  cent,  what  should  be  paid  for  it  ? 

7.  A  debt  of  $1850  is  discharged  by  two  payments  of 

f  1000  each,  at  the  end  of  one  and  two  years.  Find 
the  rate  of  interest  paid. 

8.  Reckoning  interest  at  4  per  cent,  what  annual  premium 

should  be  paid  for  30  years,  in  order  to  secure  $2000 
to  be  paid  at  the  end  of  that  time,  the  premium 
being  due  at  the  beginning  of  each  year  ? 

9.  An  annual  premium  of  $150  is  paid  to  a  life-insurance 

company  for  insuring  $5000.  If  money  is  worth  4 
per  cent,  for  how  many  years  must  the  premium  be 
paid  in  order  that  the  company  may  sustain  no  loss  ? 

10.  What   may  be  paid  for  bonds  due  in  10   years,  and 

bearing  semi-annual  coupons  of  4  per  cent  each,  in 
order  to  realize  3  per  cent  semi-annually,  if  money 
is  worth  3  per  cent  semi-annually  ? 

11.  When  money  is  worth  2  per   cent  semi-annually,  if 

bonds  having  12  years  to  run  and  bearing  semi- 
annual coupons  of  3 J  per  cent  each,  are  bought  at 
114^,  what  per  cent  is  realized  on  the  investment? 

12.  If  $126  is  paid  for  bonds  due  in  12  years,  and  yielding 

3-^  per  cent  semi-annually,  what  per  cent  is  realized 
on  the  investment,  provided  money  is  worth  2  per 
cent  semi-annually  ? 

13.  A  person  borrows  $600.25.     How  much  must  he  pay 

annually  that  the  whole  debt  may  be  discharged  in 
35  years,  allowing  simple  interest  at  4  per  cent  ? 


384  ALGEBRA. 

14.  A  perpetual  annuity  of  $100  a  year  is  sold  for  $2500. 

At  what  rate  is  the  interest  reckoned  ? 

15.  A  perpetual  annuity  of  $320,  to  begin  10  years  hence, 

is  to  be  purchased.  If  interest  is  reckoned  at  3^ 
per  cent,  what  should  be  paid  for  it  ? 

16.  A  sum  of  $10,000  is  loaned  at  4  per  cent.    At  the  end 

of  the  first  j^ear  a  payment  of  $400  is  made  ;  and  at 
the  end  of  each  following  year  a  payment  is  made 
greater  by  30  per  cent  than  the  preceding  payment. 
Find  in  how  many  years  the  debt  will  be  paid. 

17.  A  man  with  a  capital  of  $100,000  spends  every  year 

$9000.  If  the  current  rate  of  interest  is  5  per  cent, 
in  how  many  years  will  he  be  ruined  ? 

18.  Find  the  amount  of  $365  at  compound  interest  for  20 

years  at  5  per  cent. 

19.  In  how  many  years  will  $20  amount  to  $150,  at  4  per 

cent  compound  interest  ? 

20.  At  what  rate  per  cent,  compound  interest,  will  $2500 

amount  to  $3450  in  7  years  ? 

21.  If  the  population  of  a  state  increases  in  10  years  from 

2,009,000  to  2,487,000,  find  the  yearly  rate  of  in- 
crease. 

22.  The  population  of  a  State  now  is  1,918,600,  and  the 

yearly  rate  of  increase  is  2.38  per  cent.  Determine 
its  population  10  years  hence. 

23.  A  banker  borrows  a  sum  of  money  at  3^  per  cent, 

interest  payable  annually,  and  loans  the  same  at  5 
per  cent,  interest  payable  quarterly.  If  his  annual 
gain  is  $441,  determine  the  sum  borrowed. 


CHAPTER   XXV. 

Continued  Fractions. 

466.    A  FRACTION  in  the  form  of 
a 


h-\- 


''  +  7+ etc. 
is  called  a  Continued  Fraction,  though  the  term  is  commonly 
restricted  to  a  continued  fraction  that  has  1  for  each  of  its 
numerators,  as  1 

466.    Any  proper  fraction  in  its  lowest  terms  may  he  con- 
verted into  a  terminated  continued  fraction. 

Let  -  be  such  a  fraction  ; 
a 

then 


a      a  ,   c 

6      P  +  6 

(if  J)  is  the  quotient  and  c  the  remainder  of  a  ^  6) ; 
1  1  1 


and 


.  c  ,1  ,       1 


h      ^  '  b      ^  '       ^  d 
c  ^      c 


(if  q  is  the  quotient  and  d  the  remainder  of  6  -f  c), 


386  ALGEBRA. 

The  successive  steps  of  the  process  are  the  same  as  the  steps  for 
finding  tlie  G.C.M.  of  a  and  b ;  and  since  a  and  b  are  prime  to  each 
other,  a  remainder,  1,  will  at  length  be  reached,  and  the  fraction 
terminates. 

457.    The  fractions  formed  by  taking  one,  two,  three, 

of  the  quotients  jt;,  q,  r,  ,  are 

11  1 


<l      '      '  ,1 

which  simplified  are  ^^      r 

1^        q  qr^\  

and  are   called  the   first,  second,  and  third  convergents, 
respectively. 

458.    The  successive  conver gents  are  alternately  greater 
and  less  than  the  true  value  of  the  given  fraction. 

1 


Let  X  be  the  true  value  of 


V  + 


r  +  etc. 
then,  since p,  g,  r,  prepositive  integers, 

1 


P<P  + 


r  +  etc. 


-  >► ;   that  is,  -  >•  X. 

r  +  etc. 


Again,  ^  <  f/  + 


r  +  etc. 
1  1 

r  +  etc. 
< ;   that  is,  7<.x  ; 


1^  1'    ^"^^^*'  1 

P  +  -     P+-~-j  P  +  - 

</  H — 

r  +  etc,  and  so  on. 


CONTlNUtiD    FRACTIONS.  387 


459.    If  —^  — '   —   ctre  any  three  consecutive  convergents, 

and  if  nii,  11I2,  1113  are  the  quotients  that  produce  them,  then 
II3  _  niaUa  +  Ui. 
Vo        niaVa  +  Vi 

For,  if  the  first  three  quotients  are  p,  q,  r,  the  first  three  conver- 
gents  are 

P+-  P+—1 

that  IS,  -'  — ^»  ; ^-— — (11.) 

From  (i.)  it  is  seen  that  the  second  convergent  is  formed  from  the 

first  by  writing  in  it  p  +  -  for  p  ;  and  the  third  from  the  second  by 

1  ^ 

writing  q  +  -  for  q.      In  this  way,  any  convergent  may  be  formed 

from  the  preceding  convergent. 

Therefore,  —  will  be  formed  from  —  by  writing  m^  H for  1112. 

In  (ii.)  it  is  seen  that  the  third  convergent  has  its  numerator 
=  r  X  (second  nuiaerator)  +  (first  numerator)  ;  and  its  denominator 
=  r  X  (second  denominator)  -f  (first  denominator). 

Assume  that  this  law  holds  good  for  tlic  third  of  the  three  consecu- 
tive convergents 

Wo  Ml  M2  ,     ,        U2         W2M1  +  Wo 

—  '       — '       — '  so  that,     —  = ; : 

Vo  Vi  Vi  V2  7n2Vi  +  Vo 

then,  since,  by  (ii.),  —  is  formed  from  —  by  using  m2  H for  m2, 

^    '    Vs  V2  nia 


(m2  H )wi  +  Wo 


nis  {moUi  +  Up)  +  wi 


(,,,  +  _L),,  +  ,,       ^ns{m2V,  +  Vo)  +  v, 


I  mo  T 

ms 

Substitute  W2  and  V2  for  their  values  7n2Wi  +  Wo  and  niiVi  +  Vq  ; 
then  W8^m3W2  +  Wi. 

Vs       msV2  +  vi 

Therefore,  the  law  still  holds  good  ;  and  as  it  has  been  shown  to  be 
true  for  the  third  convergent,  the  law  is  general. 


Mo 

and 

—  so  that 

'  Wo" 

Ml  _  MoVi  ~  ?<iWo 
Vl                  Wo«l 

1 
WoWl' 

m2 

Mi_ 

_  M2U1  '^  M1V2 

V2 

Vl 

V1V2 
(m2Mi  +  Mo)  Ul  '^ 

Ml  (msVi  +  vo) 

388  ALGEBRA. 

460.    The  difference  between  two  consecutive  convergents 

Ui        ,  U2  .      1 

—  and  —  IS 

Vl  V2        V1V2 

The  difference  between  the  first  two  convergents 

1 g       ^  1  ■ 

P       P7  +  1       p  {pq  +  1) 

Let  the  sign  ^^  mean  "the  difference  between,"  and  assume  the 
proposition  true  for 


then 


U1U2 
(by  substituting  for  M2  and  V2  tlieir  values,  m2Mi  +  Mq  and  m2Vi  4-  vq), 

VIV2      ' 

= (by  the  assumption). 

Hence,  if  the  proposition  be  true  for  one  pair  of  consecutive  con- 
vergents, it  will  be  true  for  the  next  pair  ;  but  it  has  been  shown  to  be 
true  for  the  ^rsi  pair,  therefore  it  is  true  for  every  pair. 

Since  the  real  value  of  x  lies  between  two  consecutive  conver- 
gents, —  and  —  '  —  will  differ  from  x  by  a  quantity  less  than  —^^—^ 

Vl  V2       Ul  J        -I  J  Vl        V2 

that  is,  by  a  quantity  <" ;  so  that  the  error  in  taking  —  for  x  is 

V1U2  Vl 

<; ,  and  therefore  <  —x- 

V1V2  Vl^ 

Any  convergent,  —  ,  is  in  its  lowest  terms ;  for,  if  mi  and  Vi  had  any 

common  factor,  it  would  also  be  a  factor  of  M1V2  '^  ^2^1  (§  146) ;  that 

is,  a  factor  of  1. 


CONTINUED    FRACTIONS.  389 

46  Ir   The  successive  convergents  approach  more  and  more 
nearly  to  the  true  value  of  the  continued  fraction. 

Let  — '  — »  —  be  consecutive  convergents. 

vo      »1     »2 
Now  —  differs  from  x,  the  true  value  of  the  fraction,  only  by  using 

m2  instead  of  m^  +  — , — — -  • 
W3  +  etc. 

Let  this  complete  quotient,  which  is  always  greater  than  unity,  be 

represented  by  M. 

^,           .                    M2       wia^i  4-  Mo             Mu\  +  Mo 
Then,  smce  -  = r —  '  ^  =  17 — T 

M|  _  MUi  +  Mo        Mi  _     MqOi  ^^  MiPo    _  1 

'^  Ul  ~  MVi   +   Uo  ~  Wl  Ul  (^V»i  +  Vq)         Vi{MVi+Vo) 

Mo^      ^Uq  ^  MU]  +  Mo  _  M  {UoV\  ^  M|Po)  _  M 

Now  1  <  Jf  and  Vi  >  uq,  and  for  both  these  reasons : 

Vi  Wo 

That  is,  —  is  nearer  to  x  than  —  is. 

Vi  Vq 

462.    Any  convergent  —   is  nearer  the  true  value  x  than 
any  other  fraction  with  smaller  denominator. 

Let  T-  be  a  fraction  in  which  h<iv\. 

0 

If  -  is  one  of  the  convergents,  x~t-<  —  '>^a;.  §  461. 

If  r  is  not  one  of  the  convergents,  and  is  nearer  to  x  than  —  is, 

0  Vy 

,.       ,     ,  Ml  ,   M2      a  ^   ,  ^      ^2  ^1.         Ml 

then,  smce  x  lies  between  —  and  — ,  r  nmst  be  nearer  to  —  than  — 

Vi  Vi      0  V2  Vi 

is ;  that  is, 

a       U2   ^U\       M2           D2a~M26     .     1 
-^^^ —  <r  —  ^  —  ,  or  r- —  <  —  ; 

h  V2  Vi  W2  ^2^  ^1^2 

and  since  6  <C  ui,  this  would  require  V2a  ^  u^b  to  be  -<  1 ;  but  v^a  ^^ 
u^b  cannot  be  less  than  1,  for  a,  6,  M2,  U2  are  all  integers. 


390  ALGEBRA. 


Applications. 

(1)    Find  the  continued  fraction  equal  to  f  i,  and  also  the 
successive  convergents. 

By  following  the  process  of  finding  the  G.C.M.  of  81,  75,  the 
successive  quotients  are  found  to  be  2,  2,  2,  1,  1,  2.  Hence  the 
continued  fraction  is 


1 


2  + 


2+^ 


2  +  ^ 


1  +  -^ 


-I 

To  find  the  successive  convergents  : 

Write  the  successive  quotients  in  line,  ^  under  the  first 
quotient,  ^  under  the  second  quotient,  and  then  multiply 
each  term  by  the  quotient  above  it  and  add  the  term  to  the 
left  to  obtain  the  corresponding  term  to  the  right.     Thus, 

Quotients      =  2,  2,  2,  1,  1,  2. 
Convergents  =  f,  h  h  ts^  tV.  il- 

Note.  It  is  convenient  to  begin  to  reckon  with  f,  but  the  next 
convergent,  in  this  case  ^,  is  called  the  first  convergent. 

463.  A  quadratic  surd  may  be  expressed  in  the  form  of 
a  non-terminating  continued  fraction. 

(2)  To  express  Vs  in  the  form  of  a  continued  fraction. 

Suppose         V3  =  1  +  -  (for  1  is  the  greatest  integer  in  VS), 

then  -  =  Vs  -  1. 


CONTINUED    FRACTIONS.  391 

V3  +  1     ,  ^iC,    ,  .  ,,        .  .'  .       .   V3  +  1V 

Suppose  — - —  =  1  +  -  \^for  1  IS  the  greatest  integer  in  — h 


2  y 

1       V3+I       ^        V3-I 
then  -  =  — 1  = 


y  2  2 

2  V3+I 


y  = 


(by  §  278). 


V3-I  1 

Suppose  — '— =  2  +  -  (  as  2  is  the  greatest  integer  in  — '— )  i 

1  z  ^  .  1       ^ 

z  1 

(the  same  as  x  above). 


V3-I 
.-.  \/8  =  1  4- r,  of  which will  be  a  continu- 

1  +  2+etc.      1  +  2 

ally-repeating  series,  and  the  whole  expression  may  be  written, 

^1+2 

The  convergents  will  be  1,  2,  f ,  |,  |f,  f |,  l\,  etc. 

464.  A  continued  fraction  in  which  the  denominators 
recur  is  called  a  periodic  continued  fraction. 

465.  The  value  of  a  periodic  continued  fraction  can  be 
expressed  as  the  root  of  a  quadratic  equation. 

(3)    Find  the  surd  value  of  -      -• 


then 


Let  X  be  the  value ; 

1 


'  +  2-T-» 


2  -\-  X 
that  is,      X  =  '1         ' 

whence     x=  —l-{-  Vs. 


392  ALGEBRA. 

466.   An  exponential  equation  may  be  solved  by  con- 
tinued fractions. 

(4)    Solve  10-^  =  2,  by  continued  fractions. 

Suppose        X  =  0  +  - , 

then  10^  =  2  ; 

or  10  =  2^. 

.-.?/  =  3  +  -  (as  10  lies  between  2^  and  2*). 

Then  10  =  2^  x  2^  ; 

or 

and  2  =  (|)«. 

.•.2  =  3  +  -- 

V 

Then  2=(|)3x(|)^'; 

or 
and 

The  greatest  integer  in  v  will  be  found  to  be  9. 

1 


Hence,  x  =  0  -{■ 


3  +  -i 


^  "*"  9  +  etc. 
The  convergents  will  be  ^,  j\,  ||,  etc. 
.-.  X  =  If  =  0.3010,  nearhj. 

Exercise  125. 

1.  Find  continued   fractions    for    ||-^;    VV*  5    V5 ;    Vll  ; 

4  V6 ;  and  find  the  fifth  convergent  to  each. 

2.  Find  continued  fractions  for   2V7  i    lol  5   UU  5    Wh"  5 

and  find  the  third  convergent  to  each. 

3.  Find  continued   fractions   for   V2i  ;   V22;   V33;   ^55. 

4.  Obtain  convergents,  with  only  two  figures  in  the  denom- 

inator, that  approach  nearest  to  the  values  of  VlO ; 
Vi5;  Vl7;  VIS;   V20. 


CONTINUED    FRACTIONS.  393 

5.  Find  the  proper  fraction  which,  if  converted  into  a  con- 

tinued fraction,  will  have  quotients  1,  7,  5,  2. 

6.  Find  the  next  convergent  when  the  two  preceding  con- 

vergents  are  y\  and  ^|,  and  the  next  quotient  is  5. 

7.  If  the  pound  troy  is  the  weight  of  22.8157  inches  of 

water,  and  the  pound  avoirdupois  of  27.7274  inches, 
find  a  fraction  with  denominator  <  100  which  shall 
differ  from  their  ratio  by  <  0.0001. 

8.  The  ratio  of  the  diagonal  to  a  side  of  a  square  being 

V2,  find  a  fraction  witli  denominator  <  100  which 
shall  differ  from  their  ratio  by  <  0.0001. 

9.  The  ratio  of  the  circumference  of  a  circle  to  its  diame- 

ter being  3.14159265,  find  the  first  three  convergents, 
and  determine  to  liow  many  decimal  places  each  may 
be  depended  upon  as  agreeing  with  the  true  value. 

10.  Two  scales  whose  zero  points  coincide  have  the  dis- 

tances between  consecutive  divisions  of  the  one  to 
those  of  the  other  as  1 :  1.06577.  Find  what  divi- 
sion-points most  nearly  coincide. 

11.  Find  the  surd  values  of 

3  4.1  i.  i  1  i.  1 .  i  1  i. 

^1+6'    3  +  1  +  6'       ^2  +  3  +  4 

12.  Show  that  tlie  ratio  of  the  diagonal  of  a  cube  to  its  edge 

may  be  nearly  expressed  by  97  :  56.  Find  the  limit 
of  the  error  made  in  taking  this  ratio  for  the  true 
ratio. 

13.  Find  a  series  of  fractions  converging  to  the  ratio  of 

5  hours  48  minutes  51  seconds  to  24  hours. 

14.  Find  a  series  of  fractions  converging  to  the  ratio  of  a 

cubic  yard  to  a  cubic  meter,  if  1  cubic  yard  =  0.76453 
of  a  cubic  meter. 


CHAPTER  XXYI. 

Th?]oky    of    Limits. 

467.  When  a  quantity  is  regarded  as  having  a  fixed 
value,  it  is  called  a  Constant;  but  when  it  is  regarded, 
under  the  conditions  imposed  upon  it,  as  having  an  indefi- 
nite number  of  different  values,  it  is  called  a  Variable. 

468.  When  it  can  be  shown  that  the  value  of  a  variable, 
measured  at  a  series  of  definite  intervals,  can,  by  indefinite 
continuation  of  the  series,  be  made  to  differ  from  a  given 
constant  by  less  than  any  assigned  quantity  however  small, 
but  cannot  be  made  absolutely  equal  to  it,  the  constant  is 
called  the  Limit  of  the  variable  ;  and  the  variable  is  said 
to  approach  indefinitehj  to  its  limit. 

469.  In  order,  then,  for  a  fixed  value  to  be  the  limit  of 
a  variable  value,  it  is  necessary  and  sufficient  that  there  be 
some  diiference  between  the  variable  and  the  fixed  value, 
but  that  this  difference  may  be  made  as  small  as  ive  j^leas^. 

Consider  the  series 


1111       1 

->     2'     4'     ¥'     TG?     3^5 


The  general  formula  for  the  sum  of  7i  terms  of  this  series 
is  (§  395), 


ar" 


1-r      1- 


When  n  =  3,  the  sum  equals  r  —  j    =  2  —  -r* 

When  n  —  i,  the  sum  equals  t  —  ^  =  2  —  -• 
it  « 


THEORY    OF    LIMITS.  395 

1        --  1 

When  n=  5,  the  sum  equals  t  —  ^  =  2  —  -r^' 
i       i  16 

When  n=  (j,  the  sum  equals  7  ~  ^  =  2  —  — • 

As  n  increases,  the  sum  approaches  2  as  a  limit ;  for,  however  great 
n  becomes,  there  will  continue  to  be  some  difference  between  the 
variable  and  2,  and  this  difference,  by  increasing  n,  may  be  indefi- 
nitely diminished. 

470.  In  the  above  series  the  variable  sum  is  increasing 
towards  its  limit,  and  each  successive  value  is  less  than  its 
limit. 

A  variable  may  decrease  towards  its  limit,  and  each  suc- 
cessive value  be  greater  than  its  limit. 

A  variable,  in  approaching  its  limit,  may  be  sometimes 
greater  and  sometimes  less  than  its  limit. 

Thus,  in  the  series 

1,-i,  +i,  -i,  +xV-5V  + '   ' 

for  which  a  of  the  general  formula,  §  395,  is  1,  and  r  is  —  |. 

1  —  it         2        1 
When  n  =  3,  the  sum  equals  -  —  — r^    ~  o  +  To* 

1  '-r  2  1 

When  71  =  4,  the  sum  equals  -  —  ^      =  7;~  :n' 

2  2  3      24 

When  n  =  5,  the  sum  equals  -  —  — ^  ~  ^  "^"  73* 

1  -x"  2        1 
When  n  =  C,  the  sum  equals  -—\^      —  o~  7^-' 

2  2  3     yo 


As  n  increases,  the  sum  evidently  approaches  f  as  a  limit ;  and  for 
every  odd  number  of  terms  is  greater,  for  every  even  number  of  terms 
is  less,  than  the  limit.  Each  successive  value  of  the  variable  differs 
from  f  less  than  the  preceding  value  ;  and  though  the  successive 
values  of  the  variable  are  alternately  greater  and  less  than  the  limit, 
the  difference  between  the  variable  and  the  limit  diminishes  indefi- 
nitely by  increasing  n. 


396 


ALGEBRA. 


471.  As  a  geometrical  representation  of  the  indefinite 
approach  of  a  variable  to  its  limit, 
suppose  a  square  ABCD  inscribed 
in  a  circle  ABCD. 

Connect  the  extremities  of  each 
side  of  the  square  with  points  of 
the  circumference  equally  distant 
from  them,  namely,  the  points  E, 
F,  H,  K. 

The  length  of  the  perimeter  rep- 
resented by  the  dotted  lines  is  greater  than  that  of  the 
square,  since  two  sides  replace  each  side  of  the  square  and 
form  with  it  a  triangle  ;  and  two  sides  of  a  triangle  are 
together  greater  than  the  third  side. 

By  continually  repeating  the  process  of  doubling  the 
number  of  sides  of  each  resulting  inscribed  figure,  the 
length  of  the  perimeter  will  increase  with  the  increase  of 
the  number  of  sides ;  but  it  cannot  become  equal  to  the 
length  of  the  circumference,  for  the  perimeter  will  continue 
to  be  made  up  of  straight  lines,  each  one  of  which  is  less 
than  the  part  of  the  circumference  between  its  extremities. 
The  length  of  the  circumference  is  therefore  the  limit 
of  the  length  of  the  perimeter  as  the  number  of  sides  of 
the  inscribed  figure  is  indefinitely  increased. 

Theorems  of  Limits. 


472.  If  two  variables  are  equal  and  are  so  related  that  a 
change  in  one  produces  such  a  change  in  the  other  that  they 
continue  equal,  and  each  approaches  a  limit,  their  limits  are 
equal. 

Let  X  and  y  represent  two  equal  variables  which  increase  towards 
their  respective  limits  a  and  6,  and  which  continue  equal  in  approach- 
ing their  limits  ;  then  a  =  b. 


THEORY    OF    LIMITS.  397 

If  a  and  h  are  not  equal,  one,  as  a,  is  the  greater. 

Let  d  =  a  —  b. 

Since  x  approaches  a  indefinitely,  the  difEerence  between  a  and  x 
may  be  made  less  than  d. 

Then  a  —  «  <  d ; 

that  is,  a  —  x<Ca  —  b. 

Therefore,  »  >  6. 

Since  6  is  the  limit  of  y,  and  y  increasing, 

But  X  >  6. 

Therefore,  y  <.x. 

But  this  is  contrary  to  the  supposition  that  x  and  y  are  equal. 
Therefore,  a  and  b  cannot  be  unequal.     Hence,  a  =  b. 

When  the  variables  are  decreasing  towards  their  limits,  if  a  and  b 
are  unequal,  one,  as  6,  is  the  greater. 

Let  d  =  b  —  a. 

Since  x  approaches  a  indetiuitely,  the  difference  between  x  and  a 
may  be  made  less  than  d. 

Then  x  —  a  <  d  ; 

that  is  x  —  a<.b  —  a. 

Therefore,  x<,b. 

Since  b  is  the  limit  of  y,  and  y  decreasing, 

y>b. 

But  x<b. 

Therefore,  2/  >•  3^- 

But  this  is  contrary  to  the  supposition  that  x  and  y  are  equal. 

Therefore,  a  and  b  cannot  be  unequal.     Hence,  a=  b, 

473.  If  two  variables  have  a  fixed  ratio  and  are  so  related 
that  a  change  in  one  produces  such  a  change  in  the  other  that 
they  continue  to  have  this  ratio,  and  each  approaclies  a  limits 
their  limits  are  in  the  same  ratio. 

Let  X  and  y  represent  the  two  variables,  r  their  ratio,  a  and  b  their 
respective  limits. 

Then  x:y=r:\\  that  is,  x  =  r  X  ?/. 

By  §  472,  the  limit  of  x  =  r  X  (the  limit  of  y) ; 
that  is,  a  =  r  y.  b\  hence,  a  :  &  =  r  :  1. 


398  ALGEBRA. 

474.  The  limit  of  the  sum  of  two  or  more  variables  is 
equal  to  the  sum  of  their  resi^ective  limits. 

Let  ic,  y,   2,    .....  represent  the  variables. 

a,  6,    c,    represent  their  respective  limits, 

and      d,  d',  df\ represent  the  differences  betvreen  the  variables 

and  their  respective  limits. 

If  the  variables  are  increasing, 

X  =  a  —  (f , 

z  —  c  —  d"\  and  so  on. 
.•.X  +  2/  +  Z+  ^a^h^-c+  -(d  +  d'+d''4- ). 

It  is  required  to  show  that  d-\-  d'  -\-  d"  -{■ can  be  made  less  than 

any  assigned  quantity. 

Let  q  represent  any  assigned  quantity ;  let  (Z,  d',  and  d", be  n 

in  number,  and  d  the  greatest  of  them.     Then 
d^d'^d"^ <nd. 

Now,  since  d,  the  difference  between  x  and  its  limit  a,  may  be 
diminished  at  pleasure,  it  may  be  made  less  than  -,  so  that, 

d  <  - ,  and  therefore  nd  <;  (/. 

But  d^d'-\-d"^-  <^nd. 

Therefore,  d-V  d'  \  d"  -\-  <  g  ; 

that  is  the  difference  between  the  sum  of  the  variables  x  +  ?/  +  z  + 

and  the  sum  of  their  respective  limits  a-^h  ■\-  c-\- ,  can  be  made 

less  than  any  assigned  quantity. 

Therefore,  the  limit  ofx  +  ?/  +  2+  =  a  +  &  +  c+  

475.  The  limit  of  the  product  of  two  variables  is  equal 
to  the  product  of  their  limits. 

Let  X  and  y  represent  two  variables,  a  and  h  their  respective  limits, 
d  and  d'  the  differences  between  the  variables  and  their  limits. 
If  the  variables  are  increasing, 

x  =  a  —  d, 
y  =  b  —  d'. 
.'.  xy=ab  —  db  —  d'a  +  dd' 
=  ab—  [db  +  d'a  —  dd'). 


THEORY    OF    LIMITS.  399 

Since  d  and  d'  may  be  made  as  small  as  we  please,  and  a  and  b  are 
finite  values,  the  value  of  the  expression  db  +  d'a  —  dd'  may  be  made 
as  small  as  we  please;  that  is,  the  difference  between  the  product 
of  the  two  variables  and  the  product  of  their  limits  can  be  made  less 
than  any  assigned  quantity. 

Therefore,  the  limit  of  xy  =  ab. 

If  there  are  three  or  more  variables,  the  proposition  may 
be  proved  for  two  variables,  then  for  this  product  and  a 
third  variable,  and  so  on. 

476.  By  considering  the  variables  x,  y,  s, all  equal, 

and  the  limits  a,  b,  c, equal,  it  follows  that, 

The  limit  of  any  2^oive)'  of  a  variable  is  equal  to  that 
power  of  its  limit 

477.  The  limit  of  the  quotient  of  tivo  variables  is  equal  to 
the  quotient  of  their  limits. 

Let  X  and  y  represent  the  variables, 

a  and  b    represent  their  respective  limits, 
d  and  d'  represent  the  differences  between  the  variables  and 
their  limits. 

If  the  variables  are  increasing, 

x  =  a  —  d  and  y  =  b  —  d\ 
Therefore,  _      is  the  quotient  of  the  variables, 

also  -  is  the  quotient  of  the  limits, 

and  T  —  7 V,  is  the  difference  of  the  quotients. 

0      0  —  a 


But 


a      a  —  d  ad'  —  bd 


b      b-d'  bib-  d') 

and  since  d  and  d'  may  be  decreased  at  pleasure,  the  dividend  ad'  —  bd 
may  be  made  less  than  any  assigned  value. 

Therefore,  if  b  is  not  zero,  this  quotient  can  be  made  less  than  any 
assigned  value. 

Hence,  the  limit  of  r 7,  =  r ;  that  is,  the  limit  of  -  =  -• 

'  b  —  d'      b'  y      b 


400  ALGEBRA. 


£C"  —  a" 

478.    To  find  the  limit  of  the  quotient ,  when  x 

approaches  a  as  a  limit. 

By  division, =  x«-i  +  ax«-2  +  a2a:»-3  + a«-i. 

x  —  a 

As  X  approaches  a,  x»-i  approaches  a"-i, 
aic"-2  approaches  a»-i, 
a2x«-3  approaches  a»-i ;  and  so  on. 

Since  there  are  n  terms  in  the  right  member,  the  limit  of 

x«  — a« 

=  na"^-^  as  x  approaches  a. 

x—a  ^^ 

This  result  may  be  shown  to  be  true  whether  n  is  inte- 
gral or  fractional,  positive  or  negative. 


CONVERGENCY    OF    INFINITE    SeRIES. 

479.  A  convergent  series  is  a  series  whose  sum,  as  the 
number  of  its  terms  is  indefinitely  increased,  approaches 
some  fixed  finite  value  as  a  limit.  §  369. 

480.  The  sum  of  n  terms  of  the  geometrical  progression 
a,  ar,  ai^, has  been  shown  to  be 

When  r  >  1,  ?-"  increases  with  the  increase  of  n,  and  the 
sum  of  the  series  approaches  no  fixed  limit.  Hence,  in 
this  case  the  series  is  not  convergent. 

When  r  =  1,  the  sum  equals  %,  which  represents  no  defi- 
nite value.     Hence,  in  this  case  the  series  is  not  convergent. 

When  r  =  — 1,  the  sum  equals  0  or  a,  according  as  n  is 
an  even  or  an  odd  number.  Hence,  in  this  case  the  sum 
of  the  series  does  not  indefinitely  approach  to  a  fixed  limit 
by  increasing  n,  but  fluctuates  between  0  and  a,  and  the 
series  is  not  convergent. 


THEORY    OF    LIMITS.  401 

When  r  <  1,  r",  with  the  increase  of  n,  approaches  0 


as  a  limit,  and  the  sum  approaches  the  fixed  value 
Hence,  in  this  case  the  series  is  convergent. 


a 


1  — r 


481.  It  is  necessary  and  sufficient  for  the  convergency 
of  an  infinite  series  that  the  simi  of  the  reynaining  terms 
after  the  nth  should  approach  0  as  a  limit,  as  n  increases 
indefinitely. 

482.  If,  in  a  series  of  positive  terms,  the  sum  of  the 
remaining  terms  after  the  nth  approaches  0  as  a  limit,  it  is 
evident  that  each  separate  term  after  the  nth  must  approach 
0  as  a  limit. 

483.  If  each  term  after  the  nth  approaches  0  as  a  limit, 
it  does  not  necessarily  follow  that  the  sum  of  all  the  terms 
after  the  nth  approaches  0  as  a  limit. 

(1)  To  determine  whether  the  sum  of  the  Harmonical 
series  is  convergent : 

^111         1         1  1 

1,   -> 


2     3     4         n     7i-|-l     n-\-2 

Each  term  after  the  nth  approaches  0  as  n  increases. 
The  sum  of  n  terms  after  the  rdh  term  is 


n+ln+2n+3  2 

which  is  greater  than f- h  to  n  terms:    and  therefore 

2n       2n 

greater  than  w  X  -—  ;   that  is,  greater  than  -• 

Now,  the  first  term  is  1,  the  second  term  is  i,  the  sum  of  the  next 
two  terms  is  greater  than  i,  the  sum  of  the  succeeding  four  terms  is 
greater  than  \ ;  and  so  on.  So  that,  by  increasing  n,  the  sum  will 
approach  a  number  greater  than  any  finite  multiple  of  ^. 

Therefore,  the  series  is  divergent. 


402  ALGEBRA. 

(2)    To  determine  whether  the  following  series  is  con- 
vergent : 

l_Li_L^^ 1 i L J 1 L  Jlj i L 

^1^1X2^1X2X3^        ^|^^-l^|7^^|y^  +  l^ 

The  nth  term  of  this  series  is  evidently 


\n-  1 

The  sum  of  the  reinainiiiff  terms  is  , 1-  ; — — -  +  ; r--  + 

\n      \n+  1       |?i  +  2 


which  is 


^V        n+1       (n+l)  (n  +  2)  / 


1- 
n 


and  this  is       <,-('l  +  -4-^  +  ") 

\n\        n      n^  J 

therefore,        < ,— ( :  ) ,  for,  — —  expanded  =  1  +  -+  —  + 

[n\         j_/  '        '  ,       1      ^  n      inP- 

n 
therefore,         <  —  ( ;  I 

therefore,         <  —  (  r  ) 

\n\n  —  lj 

therefore,         < 


_n/     1 

1 

{n  —  1)  \n—\ 


But — approaches  0  as  n  increases. 

(71  —  1)  In—  1 

Therefore,  the  sum  of  the  series  is  convergent. 


Tests  of  the  Convergency  of  Series. 

484.  If  the  tei^ms  of  a  series  are  all  positive^  and  the 
limit  of  the  nth  term  is  0,  then  if  the  limit  of  the  ratio  of 
the  (n-\-l)th  term  to  the  nth  term  is  less  than  1,  the  series 
is  convergent. 

In  the  infinite  series 

ai ,  a2,  as, a„,  ctn+ij  (^n+2i ? 

let  q,  after  a  certain  value  of  n,  represent  the  limit  of  the  ratio  -^^ 
and  be  less  than  1. 


THEORY    OF    LIMITS.  403 


Also,  let  k  be  some  fixed  value  between  q  and  1.     Then 

^^<A:,     ^^<ik,     '^+-'<A:, 

a„  a„-\-\  «H+2 

.-.  a„+i  <  ka„,     ttn+o  <  k^a,!,     ctn+3  <  k^(^n- 

.'.  On+i  +  a„+2+  a«+3 <a,,  (A:+  A;- +  A:^  +  ••• 

Since  k  is  less  than  1,  :; :  =  k  +  k-  -\-  k^  +  

1  —  A: 


.-.  a„+i  +  an+2  +  a„+3  +  <a„  x  ^— ^• 

A; 
Since  an  has  a  certain  finite  value,  a,,  x     __     has  a  certain  finite 

value. 

Hence  the  infinite  series  beginning  with  a„+i  is  convergent ;  and 

by  adding  to  this  the  fixed  finite  sum  ai  +  a2  +  as  + +  Oni  the 

whole  series  is  convergent. 

Thus,  in  the  series 

i  +  U   1.1.. 


11X21X2X3  1X2X3 (n—  1) 

Om+I  _  1. 

a„        n 
Hence,  the  ratio  -^^  approaches  zero  with  the  increase  of  n,  and 

the  series  is  convergent ;  which  agrees  with  what  is  shown  in  example 
(2),  §  483. 

If  </>-l,  there  must  be  in  the  series  some  term  from  which  the 
succeeding  term  is  greater  than  the  next  preceding  term  ;  so  that  tlie 
remaining  terms  will  form  an  increasing  series,  and  therefore  the  series 
is  not  convergent. 

If  7  =  1,  this  value  gives  no  explanation  as  to  whether  the  series  is 
convergent  or  not ;  and  in  such  cases  other  tests  must  be  applied. 

If  q  approaches  1  indefinitely  as  a  limit,  then  no  fixed  vahie  k  can 
be  found  which  will  always  lie  between  q  and  1,  and  other  tests  of 
convergency  must  be  applied. 

Thus,  in  the  infinite  series 

1  +  U1  + +  i+     1     ■ 


l*-      2'-      3'-  rf      {n+  ly 

g,  the  ratio  of  the  (n  +  l)th  term  to  the  nth  term,  equals 

(  -^~- J  =  (l ZTT  )  '  which  approaches  1  as  n  increases. 


404  ALGEBRA. 

Suppose  r  positive  and  greater  than  1 ;  then  the  first  term  of  the 

2 
series  is  1.     The  sum  of  the  next  two  terms  is  less  than  —  •     The  sum 

4  2'" 

of  the  next  four  terms  is  less  than  —  •    The  sum  of  the  next  eight 
g  4'"  '^ 

terms  is  less  than  —  ;  and  so  on. 
o' 

Hence,  the  sum  of  the  series  is  less  than 

1.2,4,8,  ^1-1.1.1. 

1  +  ^+47+87+  «r<l  +  ^+4^+^+ 

which  is  evidently  convergent  when  r  is  positive  and  greater  than  1. 
If  r  is  positive  and  equal  to  1 ,  the  given  series  becomes 

l  +  i  +  i  +  i+ , 

which  is  the  Harmonical  series,  and  is  shown  in  §  483  to  be  divergent. 
If  r  is  negative,  or  less  than  1,  each  term  of  the  series  is  then 
greater  than  the  corresponding  term  in  the  Harmonical  series,  and 
hence  the  series  is  divergent. 

485.  If  the  terms  of  a  series  are  alternately  positive  and 
negative,  then  when  the  terms  continually  decrease,  and  the 
limit  of  the  nth  term  is  zero,  the  series  is  convergent. 

Consider  the  infinite  series, 

«1  —  «2  +  «3  —  «4  + =F  ««  ±  (f'n+l  H=  «^«+2  ± 

The  sum  of  the  terms  after  the  nth  term  is   * 

±  [«n  +  l  —  (a«+2  —  a«+3)  —  {dn+A  —  ^h+s)  "  ], 

which  may  be  written 

±  [ttrt+l  —  an+2  +  (ftn+S  —  a«+4)  +  (a«+5  —  «n+6)  +   1- 

Since  the  terms  are  continually  diminishing,  the  series  of  groups 
in  either  form  of  expression  are  positive,  and  therefore  the  absolute 
value  of  the  required  sum  is  seen,  from  the  first  form  of  expression, 
to  be  less  than  a„+i ;  and  from  the  second  form  of  expression,  to  be 
greater  than  a„+i  —  a„+2.  But  both  a„-|-i  and  a„+2  approach  zero 
with  the  increase  of  n ;  therefore  the  sum  of  the  series  after  the  nth 
term  approaches  zero,  and  the  series  is  convergent. 

It  will  be  seen  that  if,  in  finding  the  sum  of  an  infinite  decreasing 
series  of  which  the  terms  are  alternately  positive  and  negative,  we 
stop  at  any  term,  the  error  will  be  less  than  the  next  succeeding 
term. 


CHAPTER   XXVII. 

Indeterminate  Coefficients. 

486.    A  series,  ax -\- hx- -\- cx'^ -\- dx* -\- ,  in  which  the 

coefficients,  a,  b,  c,  d are  tinite,  may,  by  taking  x  suffi- 
ciently small,  be  made  less  than  any  assigned  quantity. 

For  if  q  be  any  assigned  quantity,  and  k  the  greatest  of  the  coeffi- 
cients a,  6,  c, ,  then 


ax  +  6x2  -I-  ca;3  -|-  <-^j;  -|-  ^^2  +  kx^  -\-  

kr 

But  kx -{- kx:^  +  kx^  + =  7-^^' 

1  —X 

(as  is  evident  by  dividing  kxhy  \  —  x). 

kx 

.:  ax  +  bx'^  -\-  cx^ -{-  <  -— — ,  if  or  be  taken  less  than  1. 

\  —X 

kx 
Hence,  if  -r- —  be  taken  less  than  a. 

1  —X 

that  is,  if  X  <  — f-j ' 

q-h  k 

then  ax  +  6x2  +  cx^  -f will  be  less  than  q. 


Theorem  of  Indeterminate  Coefficients. 

487.    If  the  two  series, 

A-{-Bx  -\-  Cx^  -\- 

and  A'-]-B'x+C'x'-\- 

are  equal  for  all  finite  values  of  x,  then 

A  =  A',     B  =  B',      C=  C\  and  so  on. 


406  ALGEBRA. 

For,  since  A  +  Bx+  Cx2+ =  A' +  B'x  +  C'x^  + , 

by  transposition,    A  —  A' =  {B' —  B)x  +  (C  —  C)x'^+  

Now,  by  taking  x  sufficiently  small,  the  right  side  of  this  equation 
can  be  made  less  than  any  assigned  value  whatever  (§  486) ;  and  there- 
fore less  than  A  —  A%  if  A  —  A'  have  any  value  whatever.  Hence, 
A  —  A'  cannot  have  any  value. 

Therefore,  A  —  A'  =  0    or    A  =  A\ 

Hence,  Bx  +  Cx^  +  Dx^  +  =  B'x  +  C'x^  +iyx^  + 

or  {B  -B')x=  {C  -  C)  x2  +  {1/  -  D)  x^  + ; 

by  dividing  by  x, 

B-B'={C'-C)x+  (2)'-D)x2+  ; 

and,  by  the  same  proof  as  for  A  —  A', 

B-B'  =  0    or    B=B\ 
In  like  manner,    C  =  C%        D  =  D',  and  so  on. 

Hence,  the  equation 

^  +  i?^+6V+ =  A'-\-B'x-\-C'x^+ 

if  true  for  all  finite  values  of  x,  is  an  identical  equation; 
that  is,  the  coefficients  of  like  powers  of  x  are  the  same. 


Applications  of  the  Theorem. 

(1)    Expand  t—j- — -:— ^  in  ascending  powers  of  x. 

Assume  ,  ^"^^'^^  =  A  +  Bx+  Cx'^  +  Dx^  + » 

1  +  X  +  x2 

then,  by  clearing  of  fractions, 

2  +  Sx  =  A  +  Bx  +  Cx^  +  Dx^  -{- 

■^  Ax-hBx'^+  Cx3+ 

+  ^x2+  Bx^+  ; 

.'.2+Sx=A-\-{B+A)x  +  {C  +  B+A)x^  +  {D  +  C  +  B)x^  + 

Therefore,  by  the  Theorem, 

A  =  2,  B-hA=S,  C  +  B  +  A  =  0,  D+C  +  B=0; 
whence,      B=  1,  c  =  —  3,  D=2,  and  so  on. 

•••  1  i"^  ^^  .2  =  2  +  x-3x2  +  2x3- 

1  +  X  +  X2 


INDETERMINATE    COEFFICIENTS.  407 

In  employing  the  method  of  Indeterminate  CoelBficients,  the  form 
of  the  given  expression  must  determine  what  powers  of  the  variable 
X  must  be  assumed.  It  is  necessary  and  sufficient  that  the  assumed 
equation,  when  simplified,  shall  have  in  the  right  member  all  the 
powers  of  x  that  are  found  in  the  left  member. 

If  any  powers  of  x  occur  in  the  right  member  that  are  not  in  the 
left  member,  the  coefficients  of  these  powers  will  be  found  to  be  0, 
and  they  will  vanish,  so  that  in  this  case  the  process  will  not  be 
vitiated ;  but  if  any  powers  of  x  occur  in  the  left  that  are  not  in  the 
right  member,  then  the  coefficients  of  these  powers  of  x  must  be  put 
equal  to  0  in  equating  the  coefficients  of  like  powers  of  x ;  and  this 
leads  to  absurd  results.    Thus,  if  it  were  assumed  in  problem  (1)  that 

2  +  .3a; 


1  4-  X  +  x'^ 


=  ^x  +  J5x2  +  Cx3  + 


there  would  be  in  the  equation  simplified  no  term  on  the  right  cor- 
responding to  2  on  the  left ;  so  that,  in  equating  the  coefficients  of  like 
powers  of  x,  2,  which  is  2x'',  would  have  to  be  put  equal  to  Ox^  ;  that 
is,  2  =  0,  a  manifest  absurdity. 

(2)  Expand  {(i  —  x)\. 

Assume  (a  —  x)^  =  ^  +  Bx  +  Cx^  +  Dx'  + 

Square a-x-  A^+2ABx  +  {2AC+R^)x-^-h (2  AD+2BC)  x^-\- 

Therefore,  by  the  Theorem, 

A2  =  a,2AB=  -\,2AC+  B^  =  0,2AD+2BC=0,  etc., 

and  ^  =  a*,  2J= -— ,,   C  =  -— ,,  2>= —- 

2  a*  8  a^  16  a^ 

Hence,  (a  —  x)*  =  a' :  —  — r  — 

2  a*      8  a^      16a^ 

(3)  Find  the  fraction  in  the  form  of  — , —. ,  which  will 

^  ^  2? -f- qx -\- rx^ 

produce,  by  executing  the  indicated  division,  the 

series  1  —  5x-\-6x^-{-Sx^  —  40a;'' — 

Assume  ^    .    ^ — t—tti,  to  be  the  fraction  required. 
1  4-  Cx  +  Dx^ 

then  .   .Vtf^T..  =  1-5x4-6x2 +  8x3- 

1  4-  Cx  4-  Dx^ 


408  ALGEBRA. 

and,  by  clearing  of  fractions, 

1  +  Bx=l-5x  +  6x^    +8x3     _ 

+  Cx-50x2  +  6Cx3+  

+  i)x2     -5i>x3+  (1.) 

...  Bii.-5+C,     6-bC  +  JJ^O,     8  +  6C-5D  =  0; 
whence   D  =  4,     C  =  2,     and    JB  =  —  3. 
l-3x 


the  fraction  is 


1  +  2x  +  4x2 


Note.  It  will  be  seen  in  the  product  (i.)  that  the  column  which 
contains  x^  may  be  obtained  by  multiplying  the  third  term  of  the 
given  series  by  1,  the  second  term  by  Cx,  and  the  first  term  by  Dx^; 
also,  that  the  column  which  contains  x^  may  be  obtained  by  multiply- 
ing the  fourth  term  of  the  given  series  by  1,  the  third  term  by  Cx,  and 
the  second  term  by  Dx^ ;  and  that  each  column  is  equal  to  0. 

A  series,  any  given  number  of  whose  consecutive  terms  are  thus 
related,  is  called  a  recurring  series;  and  the  expression,  1  +  Cx  +  Dx^, 
is  called  its  scale  of  relation. 

(4)    If  ?/  =  ax  -\-  hx^  -\-  cx^  -\- find  x  in  terms  of  y. 

Assume  x  =  Ay    -h  By^     -\- Cy^        +  ; 

But  since  y  =  ax  +  hx'^  +  cx^  +  ,  by  substituting  this  value  of 

y  in  the  equation  x  =  Ay    +  By-     +  Cy^        +  

the  result  is  x  =  Aax  +  Ahx:^  +  Acx^      + 

+  ^a'-^x2  +  2  J5a6x3  +  

+  Ca^x^     +  

BcC^  =  0,     Ac +  2  Bab  +  Ca^  -  0  ; 

.       -     r,  6      „      262- ac 

whence,  A  =  -,  B= -,  C= r 

'  "  a^  a^ 

(2  62  —  ac)  y3 


Aa- 

=  1, 

Ab-{- 

A  = 

.  1 
a' 

B=- 

a 

by^  , 

a^   ' 

Note.  Ex.  (4)  is  an  instance  of  reversion  of  series.  If  x  in  the 
given  series  is  equal  to  0,  y  will  be  equal  to  0,  and  therefore  no  term 
in  the  required  series  will  be  clear  of  y.     If,  however,  the  given  series 

is  in  the  form  of  y  =  a+  bx  +  cx"^  + ,  it  is  necessary  to  substitute 

z  for  y  —  a,  so  that  z  =  bx  +  cx^  + ,  and  to  express  x  in  terms  of  z  ; 

then  to  put  y  —  a  in  place  of  z. 

If  the  given  series  ax  +  6x2  +  cx^  + jg  ^n  infinite  series,  then  x 

must  be  less  than  1,  or  the  series  will  not  be  convergent ;  also,  y,  the 
sum  of  the  given  series,  must  be  less  than  1,  or  the  assumed  series 
Ay  +  By^  +  Cy^  + will  not  be  convergent. 


INDETKRMIXATE    COEFFICIEXTS.  409 


3^ ^ 

(5)    Resolve ——- —  into  partial  fractions. 

The  denominators  will  be  x  —  2  and  x  —  3. 

3x-7  A      ,       B 

Assume  (^3^y(^^  =  ^"^  +  ^1:3  ' 

then  3x- 7=  ^(x  — 3)  +  ^(x  — 2). 

.-.  ^  +  B  =  3  and  3  ^  +  2  J5  =  7  ; 
whence,  ^  =  1  and  B  =  2. 

3x-7  1  2 


Therefore, 


(X  -  2)  (X  -  3)       X  -  2      X  -  3 


(6)    Resolve    3         into  partial  fractions. 

Since  x^  +  1  =  (x  +  1)  (x^  —  x  +  1),  the  denominators  will  be 
x+  1  and  x2  — x+  1. 

3  A      .     Bx+  C 

Assume  „  .  ,  =  — ,— -  + 


x*  4- 1      X  +  1      x2  -  X  +  1 
then  3  =  ^  (x2  -  X  +  1)  +  (ZJx  +  C)  (X  +  1) 

=  {A  +  B)x'i+{B+C  —  A)x  +  {A  +  C)) 
whence,  3=yl4-C,     B  +  C-  A=Q,    ^  +  5  =  0; 

and  ^=1,     B=— 1,     C=2. 

3  1  X  -  2 


Therefore, 


x3  +  1      X  +  1      x2  -  X  +  1 


43.3 ^2 3^ 2 

(7)    Resolve  ^      _li\2 ^"*°  partial  fractions. 


The  denominators  may  be  x,  x^,  x  +  1,  (x  +  1) 

4x3-x2-3x-2      A  ,   B  ,      C      , 
Assume —- — -—-: =  — h  -^  H 7—,  + 


x-i  (X  4-  1)-  X      x2  '   X  +  1       (X  +  1)2 

.-.  4x3-x2-3x-2=ylx(x-f  l)2  +  7i(x+l)2  +  Cx2(x+l)+Dx2 

=  {A-\-  C)x^  +  {2A-{-  B+  C+D)x^  +  {A  +  2B)x+B 

whence,  A  +  C  =  4, 

2A  +  B+C+D=-1, 

A  +  2B=  -S, 

B=-2; 

or  B=-2,     A  =  l,     C  =  3,     i)=-4. 

^      ,         4x«-x2-3x-2      12,3  4 

Therefore,  . .    ,  , ,., =---,+ 


X2  (X  +  1)2  X        X2        X  +  1         (X  +  1)2 


410  ALGEBRA. 

In  decomposing  a  given  fraction  into  its  simplest  partial 
fractions,  it  is  important  to  determine  what  form  the 
assumed  fractions  must  have. 

Since  the  given  fraction  is  the  sum  of  the  required  par- 
tial fractions,  each  assumed  denominator  must  be  a  factor 
of  the  given  denominator ;  moreover,  all  the  factors  of  the 
given  denominator  must  be  taken  as  denominators  of  the 
assumed  fractions. 

Thus,  if  the  given  denominator  can  be  expressed  by 
x»  (x  ±  ay  (x2  +  6)«  (x2  ±ax  +  h)\ 
the  denominators  of  the  assumed  fractions  must  be 


(x±a),    (x±a)2,    ; 

(x2+&),     (X2+  6)2,     ; 

(x  ±  ax  +  6),    (x2  ±  ax  +  6)2, 


Since  the  required  partial  fractions  are  to  be  in  their 
simplest  form  incapable  of  further  decomposition,  the 
numerator  of  each  required  fraction  must  be  assumed  with 
reference  to  this  condition. 

Thus,  if  the  denominator  is  x"  or  (x  ±  a)",  the  assumed  fraction 

A  A.  Ax  +  B 

must  be  of  the  form  —  or  - — ; — —  ;   for,  if  it  had  the  form  

Ax  +  B  '^"       (^  ±  a)» '  x« 

or  ^^,  it  could  be  decomposed  into  two  fractions,  and  the  partial 

fractions  would  not  be  in  the  simplest  form  possible. 

When  all  the  monomial  factors,  and  all  the  binomial 
factors,  of  the  form  cc  ± «,  have  been  removed  from  the 
denominator  of  the  given  expression,  and  there  remains  as 
a  factor  (x^±ax-\-hY  or  (x^-\-hY,  this  is  a  quadratic  which 
cannot  be  further  resolved  ;  and  the  numerator,  therefore, 
may  contain  the  first  power  of  rr,  so  that  the  assumed  frac- 
tion must  have  the  form 

Ax-[-B  Ax-\-B 

or 


a^±iax-\-b  x'^  -\-h 


indeterminate  coefficients.  411 

Exercise  126. 
Expand  to  four  terras  in  ascending  powers  of  x  : 


2  — 3ic  2\Zx  4.  —  3X  l-x-\-x 

1                               5  — 2a;  4x  —  6x' 

5.   ^ ;; ^-:r-«'  6.   :r-7— ; «'  7. 


1  — 2ic4-3ic2  l  +  3x  — ar*  1  —  2^  +  3x2 


Revert  the  series : 
8.  y=x-\-x^-\-x^-\- 10.  y=x—lx^-\-^x^—\x'^-\-' 


9.  7/=x—2x--{-3x^— 11.  ?/— a:  +  T^,  +  ^    ^     ^+ 

-^  ^  -^         '  1X2  '  1X2X3  ' 

fit  ~f~  bx 

12.  Find  the  fractions  in  the  form  — ; ; „  whose  ex- 

.         2^-j-qx-\-rx^ 
pansions  produce  the  series  : 

l-\-3x-{-2x'  —  x^- ; 

S-\-2x-^Sx'-\-7x'  + ; 


Resolve  into  partial  fractions  : 
13. 

14. 


^^'   (2a;— l)(x— 5)       ^^'    x(x'—4.)  "■"         a;^  +  l 


7x  +  l 

16         ^-2   ^. 

19      ^*'-*- 

(x+4)(x-5) 

6 

■'■A- 

"0        ^^-* 

•     (.T-l)%T  +  2) 

(^  +  3)(x  +  4) 

5a;  — 1 

,„  ^^-^-3. 

2.r^-7.T+l. 

CHAPTER   XXVIII. 

The  Exponp:ntial  Titeorem. 

488.    To  expand  a^  in  a  series  of  ascending  powers  of  x. 
a^=\l+(a-l)\-', 
therefore,  by  the  Binomial  Theorem, 

a-=l  +  x{a-l)  +  ''  f~^'^  (a  -  If 

+       1X2X3        <•*      1)  + 

By  performing  the  indicated  operations,  this  series  will  consist  of 
powers  of  x,  and  the  whole  series  can  be  rearranged  in  ascending 
powers  of  x. 

The  coefficient  of  x  will  be  found  to  be 

(a  -  1)  -  i  (a  -  1)2  +  i  (a  -  l)^  -  ^  (a  -  1)4  + 

which  may  be  represented  by  A. 

And  if  the  coefficients  of  x-,  x^,  x*, be  represented  hy  B,  C,  D, 

the  series  may  be  written 

a^  =  1 -\-  Ax -\-  Bx^  +  Cx^ -h  Dx*  + 

Now,  B,  C,  Z), may  be  found  in  terms  of  A  ;  for  this  series  is 

true  whatever  be  the  value  of  x,  since  A,  J5,  C,  D, are  functions 

of  a,  and  therefore  wholly  independent  of  the  value  of  x.     Hence, 

a'^+v  =  I  i-  A  (x  +  ij)  +  B  {x  +  vY  +  C{x+  yf  +  D  (x  +  ?/)4  +  •- 

But       a^  +  v  =  a^x  w  =  a'/  (1  +  ^x  +  Bx'^  +  Cx^  +  Dx*  +  ). 

These  two  series  are  identically  equal,  and  therefore,  by  §  487,  the 
coefficient  of  x  in  the  first  series  is  equal  to  the  coefficient  of  x  in  the 
second  series  ;  that  is. 


THE  EXPONENTIAL  THEOREM.  413 

A+2  By  +  3Cy^  +4Dy3  + =  AoM 

=  A{l+Ay-\-  By^  +  Cy^  +  I>y*+ ) 

=  A  -\-A'^y  +  ABy^  +  ACy^-\-  

And  since  these  expressions  are  identically  equal  for  all  values  of 
y,  the  coefficients  of  ?/,  y^,  y^, in  the  one  are  equal  to  the  coeffi- 
cients of  y,  y^,  ?/3, respectively,  in  the  other. 

whence,         2B=A'^;        SC=AB;  4I)  =  AC; 

-  ^      A^  ^      AB  -r.      AC 

^^d  ^  =  T'         ^  =  -3-'  ^^-T' 

A  23.2  A  Zr^ 

so  that       a^  =  1  +  ^x  +  ~-  +  ^-^^  + 

in  which  ^  =  (a  —  1)  —  i  (a  —  l)'^  +  i  (a  —  1)3  — 

489.  If  X  is  taken  sucli  that  Ax^=l  ;  then  x  =  —>  and 
the  last  series  becomes 

J=i+i+_i^^  +  _|_+ 

The  value  of  this  series  can  easily  be  computed  to  any  degree  of 
approximation,  and  is  2.7182818 

This  constant  is  denoted,  for  shortness,  by  the  letter  e,  and  is  the 
number  whose  logarithm  is  1  in  the  Napierian  system  of  logarithms. 

In  this  system  no  base  is  assumed,  but  a  logarithm  is  defined  to  be 
such  that  the  increment  of  the  number  shall  be  the  product  of  the 
number  by  the  increment  of  the  logarithm. 

± 

490.  Since  a^  =  e,  a  =  e/,  and  A  =  logger. ;  so  that 

491.  The  preceding  proof  has  proceeded  on  the  assump- 
tion that  a  is  not  greater  than  2  ;  for  the  sum  of  the  series 

{a  —  1)  —  ^  (a  —  l)^4"i  (ff'  —  1)^  — is  not  convergent  if  a 

is  greater  than  2. 

To  extend  the  proof  to  cover  all  values  of  a,  put  b^  for  a 
in  the  theorem 

a  -l-i-{iog,a)x-\-     ^^^     +^^^^^  + 


414  ALGEBKA. 

Now,  by  taking  y  small  enough,  h  may  be  made  as  great  as  we 
please,  while  a  is  not  greater  than  2.     Then 

^^=l  +  (,og.6)..+  <>^^^^^i^|flf+ 

and  by  putting  z  for  yx^ 

Hence  the  series  is  true  for  all  values  of  a. 
492.    In  the  theorem 

a-l  +  (log,a)x-h     ^^2     +1X2X3  + 
put  e  for  a,  and  observe  that  log^e  ==  1 ;  §  294. 


then  e^=^l-\-x4- 7.  +  z 7: 7: 

^    ^1X2^1X2X3 

a  result  true  for  all  values  of  x. 


Computation  of  Logarithms. 

493.  To  eximnd  log^  (1  +  x)  in  a  series  of  ascending  pow- 
ers of  X. 

Since,  by  §491,  logea  =  A  =  {a-1)  -  i  {a-l)^  + i{a-  1)3  - , 

by  putting  1  +  x  for  a,  and  therefore  x  for  a  —  1, 

loge(l  +  x)  =  x-ia:2+^x3-^x*+ 

494.  This  series  is  called  the  Logarithmic  Series,  but  is 
not  of  a  form  for  practical  use,  as  may  be  seen  by  substi- 
tuting 10  for  X ;  which  gives 

1        11        in       10'',    103       104 

logell  =  10 --  +  —  --+ 

in  which  the  values  of  the  successive  terms  are  increasing,  and  no 
number  of  them  can  be  taken  that  will  give  an  approximate  value  of 
logjl. 

It  is  necessary,  therefore,  to  obtain  from  it  other  formulas, 
as  follow^ ; 


THE   EXPONENTIAL    THEOHEM.  415 

loge(l  +  a:)  =  x-|x2  +  ix3-ix4+ 

loge  (1  —  x)  =  —  X  —  ^ x2  —  i x3  —  1  x*  — (by  putting  —  x  for  x). 

.-.  loge  (1  +  X)  -  lOge  (1  -  X)  =  2  X  +  2  (ix3)  +  2  (i  X^)  +  

=  2(x  +  ix3  +  ix5+ ). 

But  loge  (1  +  X)  -  loge  (1  -  X)  =  loge  (f^) '  §  308. 


•••l0ge([^)  =  2(X  +  ix3+ix5+  ). 

,,.         .  m  —  n.  T^i.r       m  .      1  +  X  ^ 

In  this  series  put  — ; —  for  x,  and  therefore  —  for , 

^      m  +  n  n         1— x 

then,    log,1^  =  .2  l^^^  +  lC^Y  +  IC-^)'  + !• 

In  this  last  series  put  n  +  1  for  m,  and  therefore  _      ,  .,  for  — t""' 
n  +  1 

loge =  loge  (n  +  1)  —  logeW 

~^  i2n  +  l"^3\2n+l/  "^   "    j  ' 
...loge(n+l)  =  logen  +  2J^  +  l(^)V j; 

by  which  the  logarithm  of  any  number  can  be  obtained  from  that  of 
the  preceding  number. 

495.    To  compute  logarithms  to  the  base  e  for  1,  2,  3,  4, 
^te., 

loge  1  =0.     Hence,  by  the  last  series, 

.      log.(l  +  l)  =  log.2  =  0  +  2JJ  +  3^3+g^+^^,  +  --| 

=  0.69314718 (by  computation); 

l0ge(2+l)  =  l0ge3=l0ge2  +  2Ji+3^3+^^^,+   j 

=  1.09801228 

loge 4            =2  loge 2  §305. 

=  1. 38629430 

I0ge5  =10ge4  +  2    |J+^3+dT^+  j 

=  1.60943791 

loge  10  =l0ge5  +  l0ge2 

=  2.30268509 


416  ALGEBRA. 

496.  If  iV=any  number, 

The  multiplier — :'  by  means  of  which  every  natural  logarithm 

may  be  changed  to  the  corresponding  decimal  logarithm,  is  called  the 
Modulus  of  the  common  system,  and  is  represented  by  M.     Hence, 

^  =  2.30258509 =  <^-*^^^^^^ 

497.  Common  logarithms  may  be  obtained  from  natural 
logarithms,  as  explained  in  §  496,  or  they  may  be  computed 
directly  by  adapting  to  the  common  system  the  series 

log.(«  +  l)  =  log.«  +  2{^  +  |(^y+ }• 

Thus,     logio  (n  +  1) 

!1           1  /      1       \3      1  /      1      \5  ) 

i^r+T+HiTm)  +5(2ir+T)  + j- 

and  by  means  of  this  series  a  logarithm  in  the  common  system  may  be 
computed  from  that  of  the  preceding  number. 

498.  In  finding  the  logarithm  of  a  number  which  consists 
of  more  digits  than  are  given  in  the  tables,  it  is  assumed 
that  when  the -difference  of  two  numbers  is  small  in  com- 
parison with  either  of  them,  the  difference  of  their  logarithms 
is  proportional  to  the  difference  of  the  numbers.  The  truth 
of  this  assumption  may  be  shown  as  follows : 

Iogio(n+(Z)-Iogiow  =  logio(^^^^j     =logio(l  + -) 

=  Miog,(i  +  ^)  =  3f(^-,^.+  ^i- y 

When  d  is  very  small  in  comparison  with  n,  the  terms  of  the  right 
side,  except  the  first,  will  be  so  small  that  they  may  be  neglected  ; 

so  that  logio  {n-\-  d)  —  logio  n  =  —  » 

and  as  M  is  constant  and  n  a  given  number, 

{logio  {n  +  d)  -  logio  n}  «  d. 


CHAPTER  XXIX. 

The  Differential  Method. 

499.  If  we  have  a  series  whose  terms  proceed  according 
to  some  law,  but  are  not  immediately  equidifferentj  we 
may  find  the  difference  of  every  two  consecutive  terms,  and 
thus  form  a  series  of  differences  called  the  first  order  of 
differences.  If  the  differences  of  the  terms  of  this  new 
series  be  similarly  taken,  another  new  series  will  be  formed 
called  the  second  order  of  differences,  and  so  on. 

Let  a,  b,  c^  d, be  the  terms  of  the  series. 

Then  b  —  a,  c  —  b,  d  —  c, ,  which  may  be  denoted  by  ai,  61,  Ci, 

,  will  be  the  first  order  of  differences. 

Again,  61  —a\,  Ci  —  61,  di  —  Ci,  ,  which  may  be  denoted  by  a^, 

62,  C2,  ,  will  be  the  second  order  of  differences. 

This  process  may  be  continued  as  long  as  there  are  any  differences. 

The  given  series  and  the  successive  orders  of  differences  arranged 
in  lines  will  be : 

a        b        c        d        e       f      

dh      h  Ci       di       d      

a2      b'i       C'2       dn.       

as  63       C3        


500.    Let  it  be  required  to  express  the  {n  -\-  l)th  term  of 
the  series  a,  h,  c,  d, in  terms  of  a,  a^,  a^, 

As  b  —  a  =  au     .'.b  =  a  -\-  ai  ; 

as  61  —  «!  =  a2,     •*•  bi  =  a\  +  a2  ; 

as  b'2  —  a-i  =  aa,     .*.  62  =  «2  +  «3  ;  and  so  on. 


418  ALGEBRA. 

'In  like  manner,  c  =  6  +  61  =  a  +  2ai  +  a2, 
Ci  =  bi  +  h2  =  ai-\-  2(i2  +  as, 
C2  =  62  +  &3  =  «2  +  2as  +  a4,  and  so  on. 

Likewise,  d  =  c  +  ci  =  a  +  3ai  +  3a2  +  as,        and  so  on. 

The  coefficients,  tlierefore,  of  a,  ai,  a2,  cts? »  in  tlie  expressions 

for  6,  c,  d,  ,  are  tlie  same  as  tlie  coefficients  obtained  from  tiie 

expansion  of  tlie  expression  (a  +  6)» ;   and  since  by  the  Binomial 
Theorem  the  coefficients  of  (a  +  6)"  are 

n(n-l)       n{n-l){n-2) 
'    ^'        1x2'  1X2X3       ' 

the  {n  +  l)th  term  of  the  series  a,  6,  c,  d,  will  be 

,  ,   n(n-l)       ,  n(n-l)(n-2)       , 

501.    Let  it  be  required  to  find  the  sum  of  n  terms  of  the 
series  a,  h,  c,  d,  in  terms  of  a,  a^,  a^^ 

The  sum  of  two  terms     =  a  +  6, 

of  three  terms  =  «  +  &+€, 

of  four  terms    —  a-\-  h-\-  c  +  d^ 

and  so  on ;  so  that,  if  another  series  be  formed, 

0,  a,  a  +  6,  a  +  &  +  c, , 

the  (n  +  Vjth  term  of  this  series  will  be  the  sum  of  n  terms  of  the 
series  a,  6,  c,  d, 

But,  by  the  preceding  proof,  the  (n  +  1)^^  term  of  the  series 

0,  a,  a  +  &,  a-\-h  +  c^ 

-    ,  .  n{n  —  l)       .  n{n—l){n  —  2)       . 

°  +  '"'  +  -T^Y-°'+      1X2X3     ''^+ 

Hence,  the  sum  of  n  terms  of  the  series  a,  6,  c,  d,  is 

,   n  (n  -  1)       ,  n(7i-  1)  (n  —  2)       , 

,       ,n-\      ^  (n -  1)  (n -  2)       ^  ) 


THE    DIFFERENTIAL    METHOD.  419 

(1)  Find  the  sum  of  the  squares  of  the  first  n  natural 

numbers,  1\  2\  3^,  4^, n\ 

1      4      9      16      ri^  =  given  series. 

3      6      7        9       =  first  order  of  difEerences. 

2      2        2        =  second  order  of  differences. 

0      0       =  third  order  of  differences. 

Therefore,  a  =  1,   ai  =  3,   02  =  2,   03  =  0. 

These  values  substituted  in  the  general  formula  give 

=  ^{6  +  9n-9+2  7i2-Gn4-4} 

=  ^{2n2  +  3n+l}^^^^+Y''^^^' 
If  n  =  12,  the  sum  equals  2  X  13  X  25  =  050. 

(2)  Find  the  sum  of  the  cubes  of  the  first  n  natural  num- 

bers, 1«,  2^  3»,  4^  5», 7i». 

1      8      27      64      125    =  given  series. 

7     19      37      61      =  first  order  of  differences. 

12      18      24       =  second  order  of  differences. 

6        6       =  third  order  of  differences. 

0      =  fourth  order  of  differences. 

Hence,  a  =  1,   ai  =  7,   02  =  12,   a^  =  6,   a^  =  0. 

These  values  substituted  in  the  general  formula  give  for  the 
sum 

«  U    ,(n-l)7       (n-l)(n-2)12       (n  -  l)(n  -  2)(n  -  3)6  j 
'^j'^        2        "^  2X3  "^  2X3X4  j 

=  ^{4+  14n-14  +  8n2  — 24n  +  16  +  w3-6n2-f-lln-6} 

n^  -,   -    „,      ,      n2  (n  +  1)2       (  n  (n  +  1) 

=  -  {n^  +  2  n2  +  n}  =  — ^—: — -  =  j  — ^--r — ^ 

Now,     ^     — -  =  the  sum  of  the  first  n  numbers,  §  382 ;  hence, 

The  sum  of  the  cubes  of  the  first  n  natural  numbers  is  equal  to 
the  square  of  the  sum  of  the  numbers. 


420  ALGEBRA. 

(3)    Find  the  twelfth  term  and  the  sum  of  12  terms  of  the 
series  300,  270,  242,  216, 

300      270      242       216       =  given  series. 

—  30     —  28     —  20     =  first  order  of  differences. 

2  2        =  second  order  of  differences. 

0  =  third  order  of  differences. 

Hence,  a  =  300,  ai  =  —  30,  a^  =  2,  as  =  0,  and  n  =  12. 

These  values  substituted  in  the  general  formula  give 

Twelfth  term  =  300  -  11  X  30  +  ^^Y  ^  ^ 
=  300-330+110 
=  80. 
Sum  =  12  (300  -  J  X  30  +  -2"xT^  ^^  ) 
=  12  (300  -  165  +  36f) 
=  2060. 

Exercise  127. 

1.  Find  the  fiftieth  term  of  1,  3,  8,  20,  43, 

2.  Find  the  sum  of  the  series  4, 12,  29,  55, to  20  terms. 

3.  Find  the  twelfth  term  of  4,  11,  28,  55,  92, 

4.  Find  the  sum  of  the  series  43,  27,  14,  4,  —3, to  12 

terms. 

5.  Find  the  seventh  term  of  1,  1.235,  1.471,  1.708, 

6.  Find  the  sum  of  the  series  70,  66,  62.3,  58.9, to  15 

terms. 

7.  Find  the  eleventh  term  of  343,  337,  326,  310, 

8.  Find  the  sum  of  the  series  7  X  13,  6  X  11,  5  X  9, to 

9  terms. 

9.  Find  the  sum  of  n  terms  of  the  series  3  X  8,  6  X  11, 

9  X  14,  12  X  17, 

10.    Find  the  sum  of  n  terms  of  the  series  1,  6, 15,  28,  45, 


the  differential  method.  421 

Piles  of  Spherical  Shot. 

602.  When  the  pile  is  in  the  form  of  a  triangular  pyra- 
mid, the  summit  consists  of  a  single  shot  resting  on  three 
below ;  and  these  three  rest  on  a  course  of  six  ;  and  these 
six  on  a  course  of  ten,  and  so  on,  so  that  the  courses  will 
form  the  series, 

1,  l-f-2,  1+2  +  5,  1  +  2  +  3+4, ,1  +  2  + +  ;;. 

1       3      6       10       15       =  given  series. 

2      3      4        5       =  first  order  of  differences. 

1       1         1       =  second  order  of  differences. 

0      0        =  third  order  of  differences. 

Hence,  a  =  1,  ai  =  2,  Oa  =  1,  03  =  0. 

These  values  substituted  in  the  general  formula  give 

Sum^njl  +  -^X2+^       2X3        ^| 

=  njl  +  n— 1  + Q 

=  ^{(n  +  l)(n  +  2)} 

^n(ri+l)(n+2) 
1X2X3 

in  which  n  is  the  number  of  balls  in  the  side  of  the  bottom  course. 

503.  When  the  pile  is  in  the  form  of  a  pyramid  with  a 
square  base,  the  summit  consists  of  one  shot,  the  next 
course  consists  of  four  balls,  the  next  of  nine,  and  so  on. 
The  number  of  shot,  therefore,  is  tlie  sum  of  the  series, 

V,  2%  S%  4^ ,  7il 

XAT^.^  11      R  K(^■,    •    n  (w  +  1)  (2n  +  1) 
Which,  by  §  501,  is         ^  ^  '/^  >:^ ^' 

in  which  n  is  the  number  of  balls  in  the  side  of  the  bottom  course. 


422  ALGEBRA. 

504.  When  the  pile  has  a  base  which  is  rectangular,  but 
not  square,  the  pile  will  terminate  with  a  single  row.  Sup- 
pose p  the  number  of  shot  in  this  row ;  then  the  second 
course  will  consist  of  2{p-\-l)  shot;  the  third  course  of 
3(79  +  2);  and  the  nth  course  of  n{p)-\-n  —  l).  Hence 
the  series  will  be 

^,2^9  +  2,3^9  +  6, n{p-\-n-l). 

m 

p        2p  +  2        3p  +  6        4p+12 =  given  series. 

p+2  p  +  4:         p  +  Q        =  first  order  of  differences. 

2  2  =  second  order  of  differences. 

0  =  third  order  of  differences. 

Hence,  a  =  p,  ai  =  p  i-  2,  a2  =  2,  a^  =  0. 

These  values  substituted  in  the  general  formula  give 

Sum^n{p+^^p+2)+^---^^n7^)x2} 

=  ^  {  6p  +  3  (n  -1)  (p  +  2)  +  2  (w  -  1)  (n  -  2)} 

=  ?  (6p  +  3  np  -  Sp  +  6  71  -  6  +  2  n2  -  6  n  +  4) 

=  ^(3)ip  +  3p  +  2)i2-2) 

=  ^(n+l)(3p  +  2n-2). 

If  n'  denotes  the  number  in  the  longest  row,  then  n^  =  p  +  n—  1, 
and  therefore p  =  n'  —  n+  1;  and  the  formula  may  be  written 

^(n+l)(3n'-n+l). 

in  which  n  denotes  the  number  in  the  width,  and  n'  in  the  length,  of 
the  bottom  course. 

505.  When  the  pile  is  incomplete,  compute  the  number 
in  the  pile  as  if  complete,  then  the  number  in  that  part  of 
the  pile  that  is  lacking,  and  take  the  difference  of  the 
results. 


the  differential  method.  423 

Exercise  128. 

1.  Determine  the  number  of  shot  in  the  side  of  the  base 

of  a  triangular  pile  which  contains  286  shot. 

2.  The  number  of  shot  in  the  upper  course  of  a  square 

pile  is  1C9,  and  in  the  lowest  course  1089.     How 
many  shot  are  there  in  the  pile  ? 

3.  Find  the  number  of  shot  in  a  rectangular  pile  having 

17  shot  in  one  side  of  the  base  and  42  in  the  other. 

4.  Find  the  number  of  shot  in  five  courses  of  an  incomplete 

triangular  pile  which  has  15  in  one  side  of  the  base. 

5.  The  number  of  shot  in  a  triangular  pile  is  to  the  num- 

ber in  a  square  pile,  of  the  same  number  of  courses, 
as  22  :  41.     Find  the  number  of  shot  in  each  pile. 

6.  Find  the  number  of  shot  required  to  complete  a  rec- 

tangular pile  having  15  and  6  shot,  respectively,  in 
the  sides  of  its  upper  course. 

7.  How  many  shot  must  there  be  in  the  lowest  course  of 

a  triangular  pile  so  that  10  courses  of  the  pile,  begin- 
ning at  the  base,  may  contain  37,020  shot  ? 

8.  Find  the  number  of  shot  in  a  complete  rectangular  pile 

of  15  courses  which  has  20  shot  in  tlie  longest  side 
of  its  base. 

9.  Find  the  number  of  shot  in  the  bottom  row  of  a  square 

pile  wliich  contains  2600  more  shot  than  a  triangular 
pile  of  the  same  number  of  courses. 

10.  Find  the  number  of  shot  in  a  complete  square  pile  in 

which  the  number  of  shot  in  the  base  and  the  num- 
ber in  the  fifth  course  above  differ  by  225. 

11.  Find  the  number  of  shot  in  a  rectangular  j^ile  which 

has  600  in  the  lowest  course  and  11  in  the  top  row. 


424  ALGEBRA. 

Series  of  Separable  Terms. 

506.  It  is  evident  from  the  appearance  of  certain  series 
that  they  are  the  sums  or  the  diiferences  of  two  other 
series. 

(1)    Find  the  sum  of  the  series 

111  1 


1X2     2X3     3X4  7i(n-i-l) 

Each  term  of  this  series  may  evidently  be  expressed  in  two 
parts : 


1111  1  1 


1      22      3  ?i      n+r 

so  that  the  sum  will  be 

(i-D-a-DMhD- K^^i> 

in  which  the  second  part  of  each  term,  except  the  last,  is  can- 
celled by  the  first  part  of  the  next  succeeding  term. 

Hence,  the  sum  equals  1 — -• 

9i  +  1 

If  n  increases  without  limit,  — —  approaches  0  as  a  limit,  and 
the  sum  equals  1. 

(2)    Find  the  sum  of  the  series 

111  1 


.*.  Sum 


3X5     4X6     5X7  7i(n-\-2) 

Each  term  may  be  written, 

2\3      5/'    2V4      G/'  '  2\n      11  + 2J 

^1/1  +  1  +  1+1  + +  l_i_l_ _I_^: ^\ 

2V3      4^5      0  n      5      C  n      n+1      11+2) 

=  lfl+l 1 ^-\ 

2\3      4       n+1       ?i  +  2/ 

7  1  1 

Hence,  the  sum  equals 


24       2  (n  +  1)       2  {n  +  2) 

7 
If  n  increases  without  limit,  the  sum  equals  ~  • 


THE    DIFFEKEXTIAL    METHOD.  425 


(3)    Find  the  sura  of  the  series 
111 


3X8      6X12      9X16  3n(4.n-{-4:) 

By  multiplying  each  term  by  12  the  series  becomes 
1  1_  1  1 

1x2'   2x3'    3  X  4'  '  n{n+  1)' 

which  is  the  same  as  the  series  in  Ex.  (1). 

Hence,  when  n  increases  without  limit,  the  sum  equals  — • 

La 

(4)    Find  the  sum  of  the  series 

4  5  6  n-^S 


1X2X3      2X3X4      3X4X5  7i  (n -\- 1)  (n -\- 2) 

To  determine  whether  the  terms  of  this  series  can  each  be 
separated  into  two  parts,  assume 

w  +  3 ^        A  B 

n(n+l)(H  +  2)      n{n+\)      (n  +  l)(n4-2)* 
Reduce  the  right  member  to  a  common  denominator  ;  then 
n  +  3  ^  (^  +  B)n  +  2A 

w  (n  +  1)  (>i  +  2)      n(n+l)(n  +  2)* 
.-.  2 ^  =  3  and  A  +  B=\.  §  487. 

Whence  ^  =  t  and  B=  —  -• 

Hence,  each  term  may  be  separated  into  two  terms,  and  the 
series  will  then  become 

3/_i_+     V^  +  _l_  + ^-A—\ 

2\1X22X33X4  n(/i+l)/ 

2  V2  X  3      3  X  4  u  {n  +1)       (h+  1)  (u  +  2) / 

4      V2  X  3  ^  3  X  4  ^  ji  (ji  +  1 )  /       2  («  +  1)  («  +  2) 

4       (V2      3/^\3      4/^  ^\n      n+\)\       2(n+l)(n-f2) 

^3       11  1 


4  2      n  +  1       2  (n  +  1)  (n  +  2) 

5  3  n  +  5 

4      2(n+l)(»i+2)' 

If  w  increases  without  limit,  the  sum  equals  7- 

4 


426  ALGEBRA. 

Exercise  129. 
Sum  to  n  terms,  and  to  infinity,  the  following  series : 

11  1 

1. 


1X4     2X5     3X6 

1  1  1 

1X3X5'  2X4XG'  3X5X7'  * 
1                   1  1 

^*    2X4X6'  4X6X8'  6  X  8  X  lO' 

4  7  10 

^*    2X3X4'  3X4X5'  4X5X6'* 

1  1  1 

^'    1X2X3'  2X3X4'  3X4  X5'  * 


Interpolation  of  Series. 

507.  As  the  expansion  of  (ci  +  hy  by  the  Binomial  The- 
orem has  the  same  form  for  fractional  as  for  integral  values 
of  71,  so  the  formula 

n  (n  —  V) 
a  +  71(1^  +  "lx~2"   ^^ "^ 


may  be  extended  to  cases  in  which  ?i  is  a  fraction,  and  be 
employed  to  insert  or  interpolate  terms  in  a  series  at 
required  intervals  between  the  given  terms. 

(1)    The  cube  roots  of  27,  28, 29,  30,  are  3,  3.03659,  3.07232, 
3.10723.     Find  the  cube  root  of  27.9. 

0.03659        0.03573        0.03491  =  first  order  of  differences. 
—  0.00086     —  0.000.82         =  second  order  of  differences. 
0.00004  =  third  order  of  differences. 


Tliese  values  substituted  in  tlie  general  formula  give 

ll\/0.00004\ 
.    .      6       / 
=  3  +0.032931  +  0.0000387  +  0.00000066 
=  3.03297. 


->3-)-^(-^)(2^>T^(-fo)(-l^)(' 


THE    DIFFERENTIAL    METHOD.  427 

(2)    Given  log  127  =  2.1038,  log  128  =  2.1072,  log  129  = 
2.1106.     Find  log  127.37. 

0.0034        0.0034  =  first  order  of  differences. 

0  =  second  order  of«  differences. 

Therefore,  the  differences  of  the  second  order  will  vanish,  and 
the  required  logarithm  will  be 


=  2.1038  +  0.001358 
=  2.1052. 

(3)  The  latitude  of  the  moon  on  a  certain  Monday  at  noon 
was  1°  53'  18.9",  at  midnight  2°  27'  8.6.";  on  Tues- 
day at  noon  2°  58'  55.2",  at  midnight  3°  28'  5.8" ; 
on  Wednesday  at  noon  3°  54'  8.8".  Find  its  lati- 
tude at  9  P.M.  on  Monday. 

The  series  expressed  in  seconds,  and  the  differences,  will  be 

6798.9        8828.6         10735.2        12486.8        14048.8 

2029.7         1906.6  1750.6  1563 

—  123.1  -156  -187.6 

—  32.9  -31.6 

1.3 

As  9  hours  =  |  of  12  hours,  n  =  |. 

Also,    a  =  6798.9,    ai  =  2029.7,    a2=  — 123.1,    as  =  — 32.9 
rt4  =  1.3. 

These  values  substituted  in  the  general  formula 

,  ,   n(n-l)       ,n{n-l){n-2) 

^^  +  ^^^^+-hr2    "^+        1X2X3       ^« 

n(n-l)(n-2)(n-3) 

+     1x2x3x4     ^'^  "r 


give         6798.9+ I (2029.7) -|(-i)(^^) 


=  6798.9  +  1522.27  +  11.54  -  1.29 
^  8331.4  =  2°  18' 51,4",  • 


CHAPTER   XXX. 
The  Theory  of  jSTumbers. 

SYSTEMS    OF    NOTATION. 

508.  A  System  of  Notation  is  a  method  of  expressing 
numbers  by  means  of  a  series  of  powers  of  some  fixed 
number,  called  the  Radix,  or  Base,  of  the  scale  in  which 
the  different  numbers  are  expressed. 

509.  Integral  numbers  expressed  in  any  system  are 
polynomials  arranged  according  to  the  descending  powers 
of  the  base,  and  containing  multiples  of  these  powers 
written  in  a  condensed  form  by  omitting  the  exponents 
and  indicating  them  by  the  place  of  the  digits. 

Thus,  2384  in  the  decimal  system  means 

2  X  103  +  3  X  102  +  8  X  10  +  4  ; 
and  in  the  nonary  system  means 

2  X  93  +  3  X  9'-^  +  8  X  9  +  4. 

510.  If  r  be  any  integer,  any  integral  number  N  may  be 
expressed  in  the  form 

N=  ar""  +  br"-^  + -\-pr^  +  qr -{-  s, 

in  which  the  coefficients  a,  b,  c, ,  are  each  less  than  r. 

For,  divide  N  by  r»,  the  highest  power  of  r  contained  in  iV,  and  let 
the  quotient  be  a  with  the  remainder  N\. 

Then,  N  =  ar"  +  iVi. 

In  like  manner,  Ni  =  6r«-^  +  -A'2,     ^2  =  cr'^--  +  Ns  ;  and  so  on. 

By  continuing  this  process,  a  remainder  s  will  at  length  be  reached 
which  is  less  than  r.     So  that, 

N=  ar"  +  &r«-i  +  +  pr"^  +  qr  +  s. 


THE    THEORY    OF    NUMBERS.  429 

Some  of  the  coefficients  s,  q,  p, may  vanish,  and  each 

will  be  less  than  r;  that  is,  their  values  may  range  from 
zero  to  r — 1.  Hence,  including  zero,  the  scale  of  r  will 
contain  r  digits. 

511.    To  express  any  integral  number  N  in  the  scale  of  r. 

It  is  required  to  express  N  in  the  form  of 

ar"  4-  6r'«-i  + 4-  pr^  +  qr  +  s, 

and  to  show  how  the  digits  a,  6,  may  be  found. 

K  N=  ar"  +  6r"-J  +  +  pf^  +  qr  +  s, 

N  s 

then  —  =  civ^-^  +  6r"-2  -|-  ^  pj.  ^  ^  ^ — 

r  r 

That  is,  the  remainder  on  dividing  iV^  by  r  is  s,  the  last  digit. 

Let  Ni  =  ar"-^  +  br"-^  +  +  pr  +  q, 

then  ^  =  ar«-'^  4-  br»-^  +  +  p  +  ^• 

r  r 

That  is,  the  remainder  is  q,  the  last  but  one  of  the  digits. 

Hence,  to  express  an  integral  number  in  a  proposed  scale, 

Divide  the  number  by  the  radix,  then  the  quotient  by  the 

radix,  and  so  on ;  the  successive  remaindei's  will  be  the  siic- 

cessive  digits  beginning  from  the  units'  place. 

(1)    Express  42897   (common   scale)  in  the  senary  scale 

(scale  of  six),  and  transform  37214  from  the  octo- 

nary  scale  (scale  of  eight)  to  the  nonary  scale  (scale 

of  nine). 


) 42897 
6)7149. 
6)1191. 

..3 
..3 

(ii.)   9)37214 

0)3363 1 

9)305 6 

6)198. 

6)33. 

5. 

..3 
..0 
..3 

9)25 8 

2.  ...  3 

Arts. 

530333. 

Ans.    23861. 

In  (ii.)  the  radix  is  8  ;  and  hence  the  two  digits  on  the  left, 
37,  do  not  mean  thirtj/seven,  but  3  X  8  +  7,  or  thirtij-one,  which 
contains  9  three  times,  with  the  remainder  4. 

The  next  partial  dividend  is  4  X  8  +  2  =  34,  which  contains  9 
three  times,  with  the  remainder  7  ;  and  so  on. 


430  ALGEBRA. 

(2)    What  is  the  radix  of  the  scale  in  which  the  number 
140  (common  scale)  is  expressed  by  352  ? 

Let  r  denote  the  required  radix  ; 
then  the  value  of  352  =  8  r^  +  5  r  -|-  2. 

Hence,  3r2+5r  +  2=  140. 

Whence,  r=^Q>.   Ans. 

512.  As  in  the  decimal  scale  it  is  convenient  to  express 
quantities  less  than  the  unit  by  means  of  decimal  fractions, 
so  radix  fractions  may  be  employed  in  any  other  scale. 

Thus,  0.2341  in  the  decimal  scale  means 

10  102  ^   103  ^   104 

And  in  a  scale  whose  radix  is  r  it  means 

fjfl  a*2  ^3  1*4 

513.  Computations  are  made  with  numbers  in  any  scale, 
by  observing  that  one  unit  of  any  order  is  equal  to  the 
radix-number  of  units  of  the  next  lower  order ;  and  that  the 
radix-number  of  units  of  any  order  is  equal  to  one  unit  of 
the  next  higher  order. 

514.  The  practice  of  many  centuries  has  decided  in 
favor  of  the  decimal  system.  For,  when  the  radix  is  a 
small  number,  as  2  or  3,  large  numbers  are  expressed  in  a 
form  so  extended  that  they  are  not  easily  comprehended ; 
and  when  the  radix  is  a  large  number,  the  series  of  separate 
symbols  that  must  be  used  is  inconveniently  large. 

515.  In  some  cases  different  systems  are  employed  for 

expressing  the  same  kind  of  quantity. 

Thus,  in  measuring  time,  the  smallest  unit  is  the  second;  the  next 
higher  unit  is  the  minute,  which  is  equal  to  60  seconds ;  the  next  higher 
unit  is  the  hour,  which  is  equal  to  60  minutes ;  while  the  next  higher 
unit  is  the  day,  which  is  equal  to  24  hours, 


THE    THEORY    OE    NUMBERS.  43l 


Exercise  130. 

1.  If  6,  7,  8,  3,  2  are  the  digits  of  a  number  in  the  scale  of 

r,  beginning  from  the  right,  write  the  algebraical 
value  of  the  number. 

2.  Find  the  product  of  234  and  125  when  r  is  the  base  of 

the  scale. 

3.  In  what  scale  will  the  common  number  756  be  expressed 

by  530  ? 

4.  In  what  scale  will  540  be  the  square  of  23  ? 

5.  Show  that  1234321  will  in  any  scale  be  a  perfect  square, 

and  find  its  square  root. 

6.  In  what  scale  will  212,  1101,  1220  be  in  arithmetical 

progression  ? 

7.  Multiply  31.24  by  0.31  in  the  scale  of  5. 

8.  Find  the  least  multiplier  of  13168  which  will  make  the 

product  a  perfect  cube. 

The  Common  System  of  Notation. 

516.  A  number  which  contains  the  factor  a,  since  it  is  a 
multiple  of  a,  may  be  written  ma. 

517.  An  even  number,  since  it  contains  the  factor  2, 
may  be  written  2  m  j  and  an  odd  number  may  be  written 
2m +  1. 

518.  A  number  which  ends  in  zero,  since  it  is  a  multiple 
of  10,  may  be  written  10  m.  A  number  which  is  a  multi- 
ple of  100  may  be  written  100  m ;  and  so  on. 


432  ALGEBRA. 

519.  If  ci  prime  number  p  is  a  factor  of  a  product  ab, 
and  is  not  a  factor  a,  it  is  a  factor  of  b. 

p)  a{qx 

pqi 

n)  P  (qz 

r^  etc. 

In  finding  the  G.C.M.  of  p  and  a,  let  gi,  gg, denote  the  succes- 
sive quotients,  and  ri,  r2, the  successive  remainders. 

Then        a  —  pqi  +  ri;  p=  q^n  +  r^  ;  etc. 
By  multiplying  each  of  these  equations  by  6, 

ah  =  bpqi  +  br^ ;  bp  =  bq^Vi  +  br^, ;  etc. 
Hence,     bri  =  ab—  bpqi ; 

and  since  p  is  a  factor  of  ab  and  of  bpqi,  it  is  a  factor  of  their  differ- 
ence bri. 

Also,  since  p  is  a  factor  of  bvi,  it  is  a  factor  of  bq2ri. 

Again,  br2  =  bp—  bq<2.r\ ;  and  since  _p  is  a  factor  of  bp  and  of  bq^Xi, 
it  is  a  factor  of  their  difference  br^,. 

In  like  manner  it  may  be  shown  that  p  is  a  factor  of  br^,  br^,  br^, 
and  so  on. 

Now,  since  p  is  a  prime  number,  it  has  no  common  factor  with  a, 
except  unity  ;  and  the  last  remainder,  in  finding  the  G.C.M.  of  p  and 
a,  will  be  1. 

Hence,  p  will  be  a  factor  of  6  x  1 ;  that  is,  a  factor  of  b. 

520.  From  the  above  proposition  it  follows  that, 

A  prime  number  which  is  not  a  factor  of  either  of  two 
other  numbers  is  not  a  factor  of  their  product. 

In  other  words, 

The  product  of  two  or  more  prime  numbers  ivlll  contain  no 
prime  factor  except  themselves. 

Hence, 

A  composite  number  can  be  separated  into  only  one  set  of 
prime  factors. 

If  two  numbers  are  prirne  to  one  another,  every  power 
of  one  will  be  prime  to  every  power  of  the  othef. 


THE    THEORY    OF    NUMBERS.  433 


a 

521.  A  fraction  ~  in  its  lowest  terms  cannot  he  equal  to 

another  fraction  —  unless  6.  is  a  multiple  of  b. 

For,  if  -  =  -,  then,  by  multiplying  by  d,  —  =  c ;  and  since  c  is 

integral,  —  is  integral. 

But  6  is  prime  to  a,  and  must  therefore  be  contained  in  d  an  inte- 
gral number  of  times. 

522.  From  the  last  proposition  it  follows  that, 

A  common  fraction  in  its  lowest  terms  will  not  produce 
a  terminating  decimal  if  its  denominator  contains  any  prime 
factor  except  2  and  5. 

For  a  terminathig  decimal  is  equivalent  to  a  fraction  with  a  denom- 
inator 10".     Therefore,  a  fraction  -  in  its  lowest  terms  cannot  be 

equal  to  such  a  fraction,  unless  lO"  is  a  multiple  of  b.  But  lO",  that 
is,  2"  X  5",  contains  no  factors  besides  2  and  5,  and  hence  cannot  be 
a  multiple  of  6,  if  6  contains  any  factors  except  these. 

523.  If  a  square  number  is  resolved  into  its  prime  factors, 
the  exponent  of  each  factor  will  be  even. 

For,  if  any  number  iV  =  a''   X  &'/    X  C"  , 

]^  =  a-/'  X  b-1  X  c2'- 


524.  Conversely  :  A  number  which  has  the  exponents  of 
all  its  prime  factors  even  will  be  a  perfect  square  ;  there- 
fore, to  change  any  number  to  a  perfect  square. 

Resolve  the  number  into  its  prime  factors,  select  the  factors 
which  have  odd  exponents,  and  multiply  the  giveii  number  by 
the  product  of  these  factors. 

Thus,  to  find  the  least  number  by  which  250  must  be  multiplied  to 
make  it  a  perfect  square. 

250  =  2  X  53,  in  which  2  and  5  are  the  factors  which  have  odd 
exponents. 

Hence,  the  number  required  is  2  X  5  =  10. 


434  ALGEBRA. 


Divisibility  of  Numbers. 

625.  If  two  numbers  N  a7id  Ni,  when  divided  by  the  same 
number  a,  have  the  same  remainder  r,  their  difference  is 
divisible  by  a. 

For,  if  N  when  divided  by  a  has  a  quotient  q  and  a  remainder  r, 
then  N~qa+r. 

And  if  Ni  wlien  divided  by  a  lias  a  quotient  gi  and  a  remainder  r, 
then  jV^i  =  q^a  +  r. 

Therefore,  N  —  Ni=  {q  —  qi)  a. 

526.  If  the  differeiice  of  two  numbers  N  and  Ni  is  divisi- 
ble by  a,  then  N  and  N"i  w7ie/i  divided  by  a  «^^i7^  Aai^e  ^/ie 
sa??ie  remainder. 


For,  if 

N—Ni=  {q  —  qi)a, 

3n 

N      Ni 

Therefore, 

N             Ni 
a       ^       a       ^ 

That  is, 

N—aq  =  Ni  —  aqi. 

527.  If  two  numbers  N  and  Ni,  when  divided  by  a  given 
number  a,  have  remainders  r  and  ii,  ^Ae/i  NNi  awe?  rii  '?/;/ie?t 
divided  by  a  z^iVZ  Aai;e  ifAe  same  reynainder. 

For,  if  N=  qa-\-  r, 

and  i^i  =:  gia  +  ri, 

then  JOTi  =  gg^ia^  +  qari  +  qiar  +  rri 

=  (qqia  +  gri  +  qir)  a  +  rri. 

Therefore,  §  526,  NNi  and  rri  when  divided  by  a  will  have  the 
same  remainder. 

As  a  particular  case,  37  and  47  when  divided  by  7  have  remainders 
2  and  5,  respectively. 

Now  37  X  47  =  1739,  and  2  X  5  =  10. 

The  remainder,  when  each  of  these  two  numbers  is  divided  by  7, 
will  be  3. 


THE    THEORY    OF    NUMBERS.  435 

528.    From  the  preceding  propositions  it  follows  that : 
A  number  is  divisible  by  2,  4,  8, if  the  numbers  de- 
noted by  its  last  digit,  last  two  digits,  last  three  digits, 

are  divisible  respectively  by  2,  4,  8, 

A  number  is  divisible  by  5,  25,  125, if  the  numbers 

denoted  by  its  last  digit,  last  two  digits,  last  three  digits, 

are  divisible  respectively  by  5,  25,  125, 

If  from  a  number  the  sum  of  its  digits  is  subtracted,  the 
remainder  will  be  divisible  by  9. 

For,  if  from  a  number  expressed  in  the  form  of 

a +106+ 102c  +  103(Z+  

a+      6+       c+       d+ is  subtracted, 

the  remamder  will  be  (10  -  1)6  +  (lO"^  -  l)c  +  (lO^  -  l)d  +  , 

and  10  —  1,  102  —  1,  10^  —  1,  will  be  a  series  of  9's. 

Therefore,  the  remainder  is  divisible  by  9. 

Hence  a  number  N  may  be  expressed  in  the  form  of 
9u-{-s  (if  s  denotes  the  sum  of  its  digits); 
and  iV  will  be  divisible  by  3  if  s  is  divisible  by  3  ;  and 
also  by  9  if  s  is  divisible  by  9. 

A  number  will  be  divisible  by  11  if  the  difference 
between  the  sum  of  its  digits  in  the  even  places  and  the 
sum  of  its  digits  in  the  odd  places  is  0  or  a  multiple  of  11. 

For,  a  number  N  expressed  by  digits  (beginning  from  the  right) 
a,  6,  c,  d, may  be  put  in  the  form  of 

N=a-\-lOb+  102c+103(Z+  

.:N-a+b-c+d- =  (10+1)6+(102- l)c+(103+l)(i  + 

But  10  +  1  is  a  factor  of  10  +1,  102-1,  10^  +  1,  

Therefore,  N—a  +  b  —  c+d— is  divisible  by  10  +  1  =  11. 

Hence,  the  number  N  may  be  expressed  in  the  form  of 

lln+(a-\-c-{- )-(b-i-d+ ), 

and  will  be  a  multiple  of  11  if  (a-{-c-\- )  — (^  +  ^^4- ) 

is  0  or  a  multiple  of  11. 


CHAPTER   XXXI. 


Imaginary   Numbers. 


529.  Any  real  number,  whether  positive  or  negative, 
integral  or  fractional,  rational  or  surd,  may  be  represented 
in  magnitude  by  a  horizontal  line  extending  from  the  zero- 
point  to  the  place  it  occupies. 


I 
0 

Thus,  if  the  line  X'X  represent  a  horizontal  line  of  unlimited 
length,  and  0  be  the  zero-point,  OX  the  positive,  and  OX'  the  nega- 
tive direction,  then  all  positive  real  numbers  will  have  their  place  in 
OX,  and  all  negative  real  numbers  in  OX'. 

630.    If  the  straight  line   OA,  which  may  be  supposed 

to  represent  h  units  of 
length,  turns  around  0 
180°,  it  comes  into  the 
negative  direction  0^4'; 
-|-  h  becomes  —  h.  After  a 
further  turn  of  180°,  it 
X  comes  again  in  the  posi- 
tive direction  OA^  and  — h 
becomes  -|-  h. 

The  turn  of  180°  signi- 
fies, then,  the    multiplica- 
tion   by    the    factor    —  1. 
^'  The   double  turn   of   180° 

signifies  the  multiplication  twice  by  the  factor  — 1. 


X' 


IMAGINARY    NUMBERS.  437 

Since  —  1  is  the  product  of  V—  1  X  V—  1,  the  turn  of 
one-half  of  180°  signifies  the  multiplication  by  the  factor 

Hence,  a  definite  geometrical  meaning  may  be  assigned 
to  such  an  expression  as  h\l—l. 

Draw  YY  through  0,  perpendicular  to  X'X;  then 
h  V —  1  may  be  represented  by  h  units  counted  on  a  line 
perpendicular  to  the  original  straight  line  at  the  zero-point. 
And  the  point  B  will  be  the  place  which  this  imaginary 
number  occupies  on  the  perpendicular. 

Multiplying  again  by  the  factor  V — 1  brings  the  line 
OB  to  the  position  OA',  opposite  OAy  and  changes  +  ^  V —  1 
to  —  b.  Multiplying  a  third  time  by  the  factor  V—  1  brings 
the  line  into  the  position  0^' and  changes  — h  to  — iV — 1. 
Multiplying  by  the  factor  V—  1  a  fourth  time  brings  the 
line  into  its  original  position  OAy  and  changes  — b\J — 1  to 
-\-h,  which  corresponds  to  the  laws  governing  the  powers 
of  V^.  §  281. 

Hence,  the  direction  factors  (as  they  may  be  called),  -j- 1, 
+  V—  1,  —  1,  —  V—  1,  indicate  that  b  units  are  to  be 
counted  from  0  in  the  directions  OX,  OY,  OJC,  OY',  re- 
spectively. To  the  positive  real  number  +  b  the  negative 
real  number  —  ^  is  opposed ;  to  the  positive  imaginary 
number  +^V — 1  the  negative  imaginary  number  — by/ — 1 
is  opposed. 

631.   The  expression  V—  1  is  called  the  Imaginary  Unit, 

and  every  pure  imaginary  number  can  be  expressed  in  the 

formofzb^V— 1.  

For  convenience  the  letter  i  is  used  for  V —  1.     Hence, 

i  =V^ 
i2=(V-l}^-l, 

i3=_iV-i  =  -V^, 

i^  =  i^Xi'  =  -lX-l  =  -^l. 


438  ALGEBRA. 

Operations  with  Imaginary  Numbers. 

532.  In  order  to  add  two  imaginary  numbers,  we  begin  at 
the  place  which  the  first  number  occupies  and  count  in  the 
direction  of  the  second  number  as  many  units  as  are  equal 
to  the  absolute  value  of  the  second  number.     Therefore, 

ai  -f-  l>i  =  (a-^b)  i. 

Hence,  the  sum  of  two  imaginary  numbers  is  also  imag- 
inary. 

It  follows  that 

ai  +  bi  =  hi  +  ai. 

533.  In  order  to  find  the  difference  between  two  imag- 
inary numbers,  we  begin  at  the  place  that  the  minuend 
occupies,  and  count  in  the  direction  opposite  to  that  of  the 
subtrahend  as  many  units  as  are  equal  to  the  absolute 
value  of  the  subtrahend.     Therefore, 

ai  —  hi  =  (a  —  h)  i. 

Hence,  the  difference  of  two  unequal  imaginary  numbers 
is  also  imaginary. 

534.  To  multiply  an  imaginary  number  by  a  real  num- 
ber means  to  form  from  the  multiplicand  a  new  number, 
in  the  same  way  that  the  multiplier  is  formed  from  the 
real  unit. 

Therefore,  ai  Xb  =  obi. 

For  h  is  formed  from  the  positive  real  unit  by  taking  this  unit  h 
times  to  be  added  ;  ai  must,  therefore,  be  taken  h  times  to  be  added. 
That  is,  ai  X  6  =  ai  +  ai  +  ai  +  taken  h  times  =  ahi. 

Again,  a  X  hi  =  ahi. 

Eor  bi  is  formed  from  the  positive  real  unit  by  forming  the  imag- 
inary unit  i  from  the  positive  real  unit,  and  taking  this  imaginary 
unit  b  times  to  be  added. 


IMAGINARY    NUMBERS.  439 

Therefore,  from  the  multiplicand  a  the  corresponding  imaginary 
number  ai  must  be  formed,  and  ai  taken  b  times  to  be  added. 
Hence,     ax  bi  =  aix  b  =  abi. 

Also,      ai  X  bl  =  —  ab. 

For  ai  X  bl  =  ai~  X  b  =  —  a  X  b  =  —  ab. 

Hence,  the  product  of  an  imaginary  number  and  a  real 
number  is  imaginary ;  the  product  of  two  imaginary  num- 
bers is  real. 

It  follows,  also,  that 

aiX  b  =b   X  ai, 
and  ai  X  bi  =  bi  X  ai. 

635.  If  ab^=d  is  substituted  in  these  three  equations, 

then  a ^ 7 ,  and  -iX  b  =  di: 

b  b 

-X  bi  =  di,  -  i  X  bl^=  —  d,  or  —  t  t  X  bi  =  d. 
b  b  b 

Whence  it  follows  from  the  general  notion  of  division, 

di d  .      di d       d d  . 

J~b^'     bi~b'     bi~~V' 

Hence,  when  the  dividend  and  divisor  are  the  one  real 
and  the  other  imaginary,  the  quotient  is  imaginary ;  when 
both  are  imaginary,  the  quotient  is  real. 

From  the  second  of  these  equations  it  is  evident  that  a 
fraction  remains  unchanged  if  both  numerator  and  denomi- 
nator be  multiplied  by  i. 

636.  Since  i^  =  —  1 ,  i^  =  —  i,  i*  =  -\-l,  it  follows,  in  gen- 
eral, when  71  signifies  any  positive  whole  number, 

that  is,  (aiy  =  ai  X  ai  X  ai n  times  =  a"  I". 


440 


ALGEBRA. 


So  that  the  power  of  an  imaginary  number  is  real  or 
imaginary  according  as  the  corresponding  power  of  i  is  real 
or  imaginary. 

637.  The  sum  of  a  real  and  an  imaginary  number,  as 
a  +  bi,  is  called  a  Complex  Number,  in  distinction  from  a 
pure  imaginary  number  bi.  Two  complex  numbers  of  the 
form  a  +  bi  and  a  —  bi  are  called  Conjugate. 

638.  In  order  to  represent  the  complex  number  a  -f  bi, 

let   X'X  be  the  line  of 
^  real  numbers,  0  the  zero- 

point,  and  Y  Y'  perpen- 
dicular to  JT'X  at  0  be 
the  line  of  imaginary 
numbers.  Let  A  be  the 
place  of  the  real  num- 
ber a,  B  the  place  of  the 
imaginary  number  bi. 
Through  A  draw  a  line 
parallel  to  YY',  and  through 
B  draw  a  line  parallel  to  X'X  Then  M,  the  point  of 
intersection,  shows  the  place  of  the  complex  number  a-[-bi\ 
and  the  straight  line  OM  represents  the  absolute  value  of 
this  complex  number. 

The  point  M  can  also  be  reached  by  laying  off  a  units 
from  0  on  the  line  OX,  and  erecting  at  A,  the  last  division 
point  on  this  line,  a  perpendicular,  and  laying  off  b  units 
on  this  perpendicular. 

In  like  manner,  the  com^^lex  numbers  —  a-\-bi,  —  a  —  bi, 
a  —  bi,  correspond  respectively  to  the  points  M',  M",  M'". 

539.  If  a  and  b  have  all  values  from  —  (x  to  +  oo  the 
point  M  will  have  every  position  in  the  infinite  plane 
determined  by  X'X  and  YY'. 


M' 

B 

M 

X'.. 

A' 

\ 

A 

yb 

M" 

B' 

M 

Y' 


IMAGINARY    NUMBERS. 


441 


The  expression  a-{-blis  then  the  general  expression  for 
all  numbers.  This  expression  includes  zero  when  a=:0 
and  ^  =  0  ;  includes  all  real  numbers  when  ^  =  0  ;  all  imag- 
inary numbers  when  a  =  0 ;  all  complex  numbers  when 
a  and  b  both  differ  from  0. 


Operations  with  Complex  Numbers. 

540.  In  order  to  add  two  complex  numbers,  a  -f  bi  and 
c-\-di,  it  is  necessary  to  proceed  from  the  place  occupied 
by  the  first  number  in  the  direction  of  the  second  number 
c-{-di,  as  far  as  the  absolute  value  of  the  second  number. 
The  sum  thus  reached  consists  of  the  sum  of  the  real  and 
the  sum  of  the  imaginary  parts  of  the  two  numbers. 

Mnd  the  sum  of  (2  -|-  3  i)  +  (—  4  +  i). 

2  +  3  i  =  OM  and  —  4  +  i  =  OM'  (§  538).  If  now  we  proceed  from 
M,  the  extremity  of  OM,  in 
the  direction  of  OM'  as  far 
as  the  absolute  value  of  0M\ 
we  reach  the  point  3f". 
Hence,  OM''  =  —  2  +  4  i,  the 
sum  of  the  two  given  com- 
plex numbers. 

The  same  result  is 
reached  if  we  first  find 
the   value    of   2 +  (—4) 

=  —  2.  That  is,  if  we  count  from  0  two  real  units  to  A", 
and  add  to  this  sum  3  i  +  i  =  4  i ;  that  is,  count  four  imag- 
inary units  from  A"  on  the  perpendicular  A"M". 

In  general,  the  sum  of  two  complex  numbers  is  a  com- 
plex number. 

The  sum  of  two  conjugate  numbers  is  a  real  number. 


X'  A' 


(a  -{-  bi)  +  (a  —  bi)  =  2  a. 


442  ALGEBRA. 

.  541,  To  subtract  c-\-di  from  a -{-hi  it  is  necessary  to 
proceed  from  the  place  occupied,  by  the  minuend  in  a 
direction  opposite  to  that  of  the  subtrahend  as  far  as  the 
absolute  value  of  the  subtrahend.  The  value  thus  reached 
is  the  difference  of  the  real  and  the  difference  of  the  imagi- 
nary numbers. 

Thus,  (a  +  bi)  —  (c  +  di)  =  (a  —  c) -}- (b  —  d)  i. 

In  general,  the  difference  of  two  complex  numbers  is  a 
complex  number. 

642.    If  a -]- bi  =^ c -\- di,  then  must  a  =  c  and  b^d;  for 

otherwise  a—  c^  {d  —  b)  i,   or =  i  ;   that  is,  a  real 

ct       0 

number  equal  to  an  imaginary  number,  which  is  impos- 
sible. 

543.  If  a-\-bi  =  0,  then  must  a  =  0,  and  ^  =  0  ;  for, 
otherwise  —  -  ^  i,  which  is  impossible. 

544.  In  order  to  multiply  a  complex  number  a-^-bihj  a 
real  number  c,  the  number  a  -\-  bi  must  be  taken  c  times,  so 
that  the  real  and  the  imaginary  parts  must  be  multiplied 
by  c.     That  is, 

(a -\-  bi)  X  c  =  (a -j-  bi)  +  (a  +  bi)  -f- c  times  =^ac-{- bcL 

Thus,  to  multiply  —  2  +  i  by  3 :    Take  0A  =  —2  on  OX',  and 

erect  at  A  the  perpendicular  AM=\.     Then  0M=  —  2  +  i;  and, 

by  taking  OM  three  times,   the   result 

is    OM'  =  —  6-\ 3 i   the   product   of 

(-2  +  i)by3. 


545.  In  order  to  multiply  a-\-bi 

by  c  -|-  di  it  is  necessary  to  obtain 

a  number  from  a  -\-  bi  in  the  same 

way  that  c  +  di  is  obtained  from 

the  positive  real  unit.     It  is  necessary,  therefore,  to  take 

a  +  bi  itself  c  times,  then  to  take  a-\-bi  in  the  imaginary 


IMAGINARY    NUMBERS. 


443 


direction  d  times,  and  join  the  two  results  in  one  sum. 
The  result  corresponds  to  the  algebraic  product  of  two 
factors. 


Find  the  product  of  (3  —  I)  (2  +  3 1). 

Find  OM  =  3  —  i.     It  is  required  to  find  a  product  from  2  +  3i  in 
the  same  way  that  OM  is  ob- 
tained from  3—1. 

With  OM  as  a  new  positive 
unit,  lay  off  OM'  -  2  OM.  At 
M'  erect  a  perpendicular,  and 
lay  off  from  M\  on  this  perpen- 
dicular, WM"  =  3  OM.  Then 
(3-i)  (2+3i)=  0M''=d+7i. 

If  2  +  3  i  is  the  multiplier, 
the  new  positive  unit  will  be 
0M''\ 

The  same  result  is  ob- 
tained if  the  product  is 
found  by  the  algebraic 
method  of  multiplication. 
Thus, 

(3  — i)  (2  +  3i)==6-2i  +  9i  +  3  =  9  +  7i. 

In  general,  the  product  of  two  complex  numbers  is  a 
complex  number. 

The  product  of  two  conjugate  numbers  is  a  real  number. 


] 

^ 

M 

// 

/-» 

M 

in            y^              T 

TT 

^<. 

^T--^^^^^ 

— 

Thus, 


{a-\-hl)  (a  —  bi)  =  a--{-b' 


646.    Since  (a  +  hi)  X  c  =  ac  +  bci,  it  follows  that 

ac  4-  hci  ,   , . 

=  a  -}-  hi. 

c 

That  is,  a  complex  number  is  divided  by  a  real  number 
by  dividing  both  the  real  and  imaginary  parts  by  this 
number. 


444  ALGEBRA. 

.      .      c(a-\-bi)       c 

That  is,  a  quotient  is  unchanged  if  both  the  dividend 
and  divisor  are  multiplied  by  the  same  complex  number. 

547.  In  order  to  divide  one  complex  number  by  another, 
it  is  only  necessary  to  multiply  the  dividend  and  divisor 
by  the  conjugate  of  the  divisor,  and  the  result  is  a  complex 
number  to  be  divided  by  a  real  number.     Thus, 

a-{-bi («  +  bi)  (c  —  di) (ac  -f-  bd)  -\-  {be  —  ad)  i 

c-{-di  ~  (c-^d{)(c  —  di)  ~  c2  +  ^2 

ac-\-bd       (be  —  ad)  i 

~  e'-i-d'  "^      e'-\-d' 

The  quotient  of  two  complex  numbers  is  a  complex 
number.     Thus, 

S-\-i  _   (S-\-i)(2  —  5i)  _  11  — 13f 

2  +  5i~"(2  +  5i)(2  — 5i)"~       29 

648.  The  power  of  a  complex  number  is  a  complex 
number.     Thus, 

(a  +  bi)''=  (a  +  bi)  (a  +  bi)  =  (a^  —  b^)-i-2 abi, 
la  +  b{)^=  (a  +  bi)^  (a  +  bi) 

=  (a^  —  3  ab^)  +  (3  a^b  —  b^)  i,  and  so  on. 

549.  The  meaning  of  an  imaginary  exponent  is  deter- 
mined by  subjecting  it  to  the  same  operations  as  if  it  were 
a  real  exponent. 

It  follows  that  such  an  expression  as  A^^^^',  where  A;  is  a 
real  number  and  a-\-bi  Si,  complex  exponent,  may  be  simpli- 
fied by  resolving  it  into  two  factors,  one  of  which  is  a  real 
number,  and  the  other  an  imaginary  power  of  e  (e  being 
used  as  in  §  489). 

By  the  ordinary  rules  for  exponents, 


IMAGINARY    NUMBERS.  445 

Put  li'  =  e'%  u  denoting  log  [A:*]  in  the  natural  system ; 
then 

The  value  of  e"'  is  (§  492) 

,272  /i/"j "  ^/^/i  ^  -j/^j  ^ 


This  value  (§§  531  and  536)  may  be  resolved  into  two 
series,  as  follows : 

^"■=i-^+g-^+ +\"-^+^- )• 

The  series 

u^      u*      u^ 
i2"^[4~JG"^ 

is  a  function  of  u  called  Cosine  of  w,  and  is  written  cos  u. 
The  series 

"-11+^- 

is  a  function  of  u  called  Sine  of  u,  and  is  written  sin  u. 

550.    Hence,  the  value  of  e"'  may  be  Avritten 

e'"  =  cos  u  -\-  i  sin  u.  (1) 

If,  in  the  development  of  e"',  u  be  changed  to  —  ?/,  cos  u 
will  remain  unchanged,  because  its  terms  contain  only  even 
powers  of  u ;  but  sin  u  will  become  —  sin  ^i,  because  its 
terms  contain  only  odd  powers  of  u.     Therefore, 

g-«*  __  pQg  ^  —  I  g|j^  ^  ^2) 

The  product  of  equations  (1)  and  (2)  is 
^       e"*  X  e-"'  =  (cos  uf  —  P  (sin  uf, 

that  is,  1  =  (cos  uy  —  P  (sin  uy,  "        (§  255) 

or  1  =  (cos  uy  +  (sin  uy.  (§  531) 


446  ALGEBRA. 

It  is  customary  to  omit  the  parentheses,  and  to  write 
this  equation  as  follows  : 

cos^  u  +  sin^  u=^l.  (3) 

551.   If  nu  is  substituted  for  w,  and  if  e""'  is  developed  as 
e"*  has  been  developed,  the  result  will  obviously  be 

But  equation  (1)  raised  to  the  nth  power  becomes 

gnwi  __  ^(.Qg  ^  _|_  {  gin  uy. 

Therefore, 

cos  nu-^i  sin  nu  =  (cos  u-\-i  sin  uy.  (4) 

If,  in  this  equation  n=^2,  the  resulting  equation  is 
cos  2u-\-i  sin  2  w  =  cos^  l^  —  sin^  u-\-2i  sin  w  cos  u] 
whence  it  follows,  from  §  542,  that, 

cos  2  ^^  =  cos^  u  —  sin^  u,  (5) 

sin  2  ?^  =  2  sin  u  cos  u.  (6) 

By  substituting  x,  y,  x-{-y,  successively,  for  u  in  equa- 
tion (1), 

gx»       =r  cos  a;  H-  i  sin  x,  (7) 

ev'       =  cos  ?/  +  i  sin  y,  (8) 

g(x+j/)i  =  cos  (cc  +  ?/)  +  i  sin  (ic  +  ?/) .  (9) 

The  product  of  equations  (7)  and  (8)  is 
g(x+j/)i  __  gQg  y.  (>Qg  2^ gijj  ^  gij^  y 

+  i  (sin  cc  cos  ?/  +  cos  x  sin  ?/). 
Therefore, 

cos  (x-\-y)-{-  i  sin  (x  +  ?/)  =  cos  cc  cos  y  —  sin  x  sin  ?/ 

+  i  (sin  a:;  cos  y  -\-  cos  a;  sin  y)  j 
whence  (§  542), 

cos  (x  -\-y)  =  cos  a;  cos  y  —  sin  x  sin  y,  (10) 

sin  (ir  +  //)  =  sin  x  cos  y  +  cos  x  sin  y.  (11) 


CHAPTER  XXXII. 

Loci    of    Equations. 

552.  It  is  possible  to  represent  equations  by  diagrams, 
some  of  which  are  regular  geometrical  figures. 

For  this  purpose,  the  lines  X'X  and  YY'  are  drawn 
perpendicular  to  each  other,  intersecting  at  the  point  0. 
These  lines  may  be  of  any  length  in  drawing  the  diagram 
of  an  equation. 


H 


X'X. and  YY'  are  called  the  Axes  of  Reference;  X'X 
is  the  Axis  of  Abscissas,  and  YY'  the  Axis  of  Ordinates. 

In  order  to  determine  the  position  of  the  point  A,  with 
reference  to  these  axes,  AE  is  drawn  perpendicular  to  X'X, 
and  is  called  the  Ordinate  of  the  point  A.  The  line  OE 
from  the  point  0  to  the  foot  of  the  ordinate  is  the  Abscissa 
of  A,  OE  and  AE  are  the  Co-ordinates  of  A.  What  are 
the  co-ordinates  of  B,  C,  and  D  ? 


448 


ALGEBRA. 


The  abscissa  of  a  point  is  usually  designated  by  the  let- 
ter £c,  and  its  ordinate  by  y.  For  the  point  J,  a:  =  OE^ 
and  y=^AE. 

0  is  the  Origin  of  Co-ordinates,  or  simply  the  Origin. 

553.  In  representing  equations  whose  roots  are  arith- 
metical numbers,  numerical  values  are  assigned  to  the 
co-ordinates.  All  ordinates  ahoiw  X'X  are  considered  posi- 
tive, and  all  below,  negative.  Abscissas  drawn  to  the  right 
of  YY'  are  positive,  and  those  to  the  left  are  negative. 

Y 


-f— <- 


E 


-I— t- 


Equal  lengths  of  any  convenient  magnitude  are  measured 
on  the  axes  from  0,  and  each  length  represents  a  unit.  To 
locate  the  point  whose  abscissa  is  4  and  whose  ordinate  is 
7,  a  distance  of  4  units  is  measured  on  X'X  from  0  to  the 
right,  and  at  the  point  reached  a  perpendicular  7  units 
long  is  erected.  A,  the  extremity  of  this  perpendicular,  is 
the  required  point. 

The  co-ordinates  of  B  are  x  =  —  6,  and  y  =  9  ;  of  (7, 
x  =  —  3,  and  y=  —  6  ;  of  J),  x^S,  and  y  =  —  8. 


LOCI    OF    EQUATIONS. 


449 


554.  The  equation  2x-\-i/  =  3  may  be  satisfied  by  an 
infinite  number  of  corresponding  values  of  x  and  y,  By 
changing  the  equation  to  the  form  y  =  3—2x,  the  follow- 
ing table  is  readily  computed  : 


If  x  =  l, 

y=    1- 

If  x  =  -l, 

y=  5. 

"  x  =  2, 

y  =  -\. 

"  x  =  —  2, 

y=  7. 

"  x  =  S, 

y  =  -3. 

"  x  =  —  3, 

y=  9. 

''  ic  =  4, 

?/  =  — 5. 

ii   x  =  —  4:, 

y=ll. 

<'  x  =  5, 

y  =  -7. 

'i  x  =  —  5, 

2/ =  13. 

This  table  contains  the  co-ordinates  of  ten  points, 
the  points  are  lo- 
cated   with    refer- 
ence  to   the   axes, 


If 


\ 


\ 


X' 


\ 


:\ 


I  \ 


and  the  line  MN 
is  drawn  through 
them,  the  line  MN 
is  the  Locus  of  the 
given  equation. 

If  MN  be  pro- 
longed, the  value 
of  y  which  cor- 
responds to  any 
given  value  of  x 
may  be  found  by 
laying  off  an  ab- 
scissa equal  to  the 
given  value  of  ic, 
erecting  at  its  ex- 
tremity an  ordinate 
terminated  by  MN,  n 

and  measuring  the  length  of  the  ord.  .ate. 

If  ic  =  0,  the  diagram  shows  tha",  y  =  3;  and  if  a^  =  1.5, 
y  =  0.     It  x  =  —  3.5,  what  is  the  v   iue  of  ?/ ? 


450 


ALGEBRA. 


565.  In  plotting  any  equation  containing  x  and  y,  we 
assume  values  of  x,  compute  the  corresponding  values  of  y, 
locate  the  points  whose  co-ordinates  are  thus  found,  and 
draw  the  locus  through  them.  When  the  equation  con- 
tains only  the  first  powers  of  x  and  y,  and  is  of  the  first 
degree,  the  locus  is  always  a  straight  line.  In  this  case  it 
is  necessary  to  locate  only  two  points  of  the  line ;  the  locus 
may  then  be  drawn  through  these  points. 


556.   Let  it  be  required  to  draw  the  locus  of 
^2  + 7/2  =.72.25. 

y- 


^12.26 -x\ 


From  this  equation,  by  assuming  values  of  x  and  com- 
puting the  values  of  y,  the  following  table  may  be  formed  : 


If  X  is 

2/ is 

±1 

±8.44  + 

±2 

±8.26  + 

±3 

±7.95  + 

±4 

±7.50 

±5 

±6.87  + 

±6 

±6.02  + 

dz7 

±4.82  + 

±8 

±2.87  + 

If  a  curve  is  drawn 
through   the    extremi- 
ties of  the  ordinates, 
the  locus   is  found  to 
be  a  circle  whose  cr^tre  is  at  the  origin.     The  diagram 
shows  that  x  does  n      reach  the  value  9.     If  this  value 
be  substituted  in  the  e    lation,  y  will  be  imaginary. 


LOCI    OF    EQUATIONS. 

557.   Construct  the  locus  of  9^2  + 25/ =  900. 


451 


y  =  ±^  V900  — 9^2. 
From  this  equation  the  following  table  may  be  formed : 


If  X  is 

y  is 

±1 

d=5.96  + 

±2 

±5.87  + 

±3 

±5.72  + 

±4 

±5.49  + 

±5 

±5.19  + 

±6 

±4.80 

±7 

±4.28  + 

±8 

±3.60 

±9 

±2.61  + 

±10 

0.00 

0 

±6. 

y^ 

/; 

1    1    1 

r     1     1 

,-r" 

-^ 

' 

_- 

^ 

-X 

— i — 1-    j  ■ 

^    !    > 

^  '    ! 

^^^ 

•■^ 

._ 

_ 

■ 

_ 

.ii-^- 

y 

y 

1 

The  locus  is  an  ellipse. 
668.   Construct  the  locus  oiif^  5x. 
In  this  case  y  =  ±  ^/5x. 


Uxis 

yis 

If  a;  is 

y  is 

1 

±2.23  + 

6 

±5.47  + 

2 

±3.16  + 

7 

±5.91    - 

3 

±3.87  + 

8 

±6.32  + 

4 

±4.47  + 

9 

±6.70  + 

5 

±5.00 

10 

±7.07  + 

For  negative  values  of  x  the 
ordinates  are  imaginar}^  The 
curve  is  a  parabola;    and  as  x 

and  y  may  have  infinite  values,  the  curve  will  extend  to  an 
infinite  distance. 


.^-T 


452 


559. 


■ 

j 

>■; 

X 

/ 

/. 

1 
1  ' 

i: 

; 

ALGEBRA. 

le  locus  of  y=^x^- 

-x^-^x- 

-5. 

If  X  is             y  is 

Ifxis 

y  is 

0.5         -4.625 

2.5 

+  6.875 

1.0         -4.000 

0.0 

-  5.000 

1.5         -2.375 

-0.5 

-  5.875 

2.0         + 1-000 

-1.5 

-12.125 

rr 


This  locus  is  not  a  regular  geometri- 
cal figure. 

Computation  shows  that,  if  y  =  0, 
«  =  1.88  + .  Does  this  result  agree  with 
the  figure  ?  What  is  x  when  ?/  is  +  5  ? 
When  2/  is  + 10  ?     When  'y  is  -  15  ? 


560.  When  two  simultaneous  equations  containing  x  and 
y  are  given,  the  values  of  x  and  y  which  satisfy  both  are 
shown  by  constructing  the  loci  of  both  equations  with  ref- 
erence to  the  same  axes.     In  the  equations 

3x  -\-  7//  _  ^       2x 


—  7       and 


y  =  -(2  +  Sx),      (1); 


y 


and 


y 


=  2x 


19. 


(2) 


A\^\    I    I 


+  111 


MM. 


/ 
/ 


AB  is  the  locus  of  equation  (1),  and  CD  is  the  locus  of 

equation  (2).  The  values 
of  the  co-ordinates  of  any 
point  on  AB  satisfy  (1),  and 
those  of  any  point  on  CD 
satisfy  (2).  Since  the  point 
E  lies  on  both  AB  and  CD, 
its  co-ordinates  satisfy  both 
equations.  By  measurement, 
they  are  found  to  be  x  =  7, 
and  y  =  —  5.  The  correct- 
ness of  these  values  is  easily  proved  by  substituting  them 


E/ 


B 


LOCI    OF    EQUATIONS. 


453 


in  the  given  equations.     Errors  in  drawing  the  diagram 
will  of  course  alfect  the  accuracy  of  the  results. 

661.   In  the  equations 

0^  +  2/^  =  65,         (1) 
x-y  =11,         (2) 

the  circle  ABCD  is  tlie 
locus  of  equation  (1),  and 
the  line  MN  is  the  locus 
of  equation  (2).  The  co- 
ordinates of  A  and  B  sat- 
isfy both  equations,  since 
A  and  B  lie  on  both  loci. 
The  co-ordinates  of  A  are 
x=-l,  y^  —  4  ;  the  co-ordinates  of  B  are  x  =  4,  y  =  —  7. 

Since  the  square  on  the  hypothenuse  of  a  right  triangle 
equals  the  sum  of  the  squares  on  the  other  two  sides, 
inspection  of  the  diagram  shows  that  (if  r  is  the  radius 
of  the  circle), 

r'=x'-\-,f. 

Hence,  the  locus  of  any  equation  of  the  form  x^-{-if^=7^ 
is  a  circle  whose  radius  is  the  square  root  of  the  right-hand 
member.     In  the  present  example 

^=±  V 65  =  ±8.06 


/""" 

\ 

\ 

.-'/if 

/ 

If 


Exercise  131. 

Solve  the  following  equations  by  constructing  their  loci : 

1.  2x  +  3?/  =  8|  3.  2a;  — %  =  11    ^ 
3ir  +  7y  =  7J  3.T-12y  =  15j 

2.  3a-  — 5//  =  2    ^  4.  4.x  —  2y  =  20    ^ 

^x^=9y  J 


la-- 5//  =  2    ^ 


454 


ALGEBRA. 


5. 

2x-^y  = 

^x^2y  = 

4   ^ 
32/ 

6. 

^x  —  hy^z] 

7. 

3ic  — 47/  =  7   / 

8. 

3a;  — 4?/  =  — 5^ 
/^x  —  hy  =  \      J 

9. 

x-2y  =  ^^ 
2x-y  =  5) 

10. 

§-1  =  5^ 

X  ■    y 

1-5  =  6 

X       y         J 

■ 

11. 

X      y 
§-?  =  4 

► 

12.  a;2  +  ?/2^-^04. 
a:  +  7/=:12      ; 

13.  £c  — ?/  =  10      ^ 
0^^  +  7/^-178; 

14.  0^?/  — 12  =  0 


17. 


-22/-5 ; 

=  36      J 


15.    a;  +  ?/  =  13 
a;2/: 


16.    3z/-  — 4x2  =  12^ 

=  -io| 


2^  +  y: 
4 


5  +  y      12  +  a; 
2x  +  5?/  =  35 


a;      3y~"27 
11 

72  J 


-  +  - 
4cc      y 


562.  When  a  single  equation  is  given  containing  only 
one  unknown  number,  all  the  terms  must  be  transposed 
to  the  first  member.     Thus, 

a;2  +  cc  — 15.75  =  0; 
or,  if?/  =  0,  x^-{-x  — 15.75=2/. 

Assuming  values  of  cc,  we  compute  the  corresponding 
values  of  y,  and  construct  the  locus.  Now,  any  value  of  x 
which  makes  2/  =  0  satisfies  the  equation,  and  is  a  root ; 
hence,  any  abscissa  whose  corresponding  ordinate  is  zero 
represents  a  root.  The  roots,  therefore,  may  be  found  by 
measuring  the  abscissas  of  the  points  where  the  locus  meets 
X'Xf  for  at  these  points  y  =  0. 


LOCI    OF    EQUATIONS. 


455 


From  the  given  equation  the  following  table 
formed  : 


may 


be 


X  is 

y  is 

If  X  is 

y  is 

0 

-15.75 

-1 

-15.75 

1 

-13.75 

_2 

-13.75 

2 

-9.74 

-3 

—  9.75 

3 

-3.75 

J. 

—  3.75 

4 

+  4.25 

-5 

+  4.25 

The  table  shows  that  one  root 
is  between  3  and  4  (since  y 
changes  from  —  to  +,  and  there- 
fore passes  through  zero);  and, 
for  a  like  reason,  the  other  is 
between  —  4  and  —  5.  The  dia- 
gram shows  that  the  roots  are 
3.5  and -4.5. 


M^ 


663.   Construct  the  locus  of  x^  —  5a;  —  3  =  0. 
From  this  equation  the  following  table  may  be  formed : 

Y 


X 

y 

X 

y 

0 

-3 

-1 

+  1 

1 

-7 

2 

-1 

2 

—  o 

-3 

-15 

3 

+  9 

To  plot  this  locus  accurately,  the 
values  of  y  must  be  computed  for 
ic  =  — 1.1,  —1.2,  —1.3,  etc. 

The  necessary  portions  of  the  lo- 
cus are  given  in  the  diagram,  and 
the   roots    are    found    to  be   2.5-, 


—  0.6 +  ,  and 
and— 1.83  + . 


1.8 +  ;  nearer  values  are  2.49  +  , —0.65  + , 


456 


ALGEBRA. 


An  equation  of  any  degree  may  be  thus  plotted,  and  the 
locus  will  be  found  to  cross  the  axis  X^X  as  many  times  as 
the  number  of  real  roots  in  the  equation. 

564.    When  an  equation  has  no  real  roots,  the  locus  does 
not  meet  XX. 
y\  \  In  the  equation 

/  a;2— 6x  +  13  =  0, 

\ 

\      /  both  of  whose  roots  are  imaginary, 

the  locus,  at  its  nearest  approach,  is 

4  units  distant  from  X'X. 


XU- 


565.   The    locus    of    an    equation 
whose  roots  are  equal  touches  X'X, 
but  does  not  intersect  it. 
The  equation 

a^^-f  4a^  +  4=0 

has  the  roots  —  2  and  —  2,  and 
its  locus  is  shown  in  the  mar- 
gin. 


aX 


Y' 


566.  Loci  are  chiefly  valued 
as  illustrations  of  the  properties 
of  equations. 


Exercise  132. 


Construct  the  loci  of  the  following  equations  : 

1.' :^2_^3x  — 10  =  0.  3.    :^^  — 20.^2  +  64  =  0. 

2.    x^  — 2a;2  +  l  =  0.  4.    a;^  — 4.'r  +  10  =  0. 

5.    ic^  — 5x^+4  =  0. 


CHAPTER   XXXIII. 

Equations  in  G-eneral. 

667.  Every  higher  equation  of  which  we  shall  treat  can 
be  reduced  to  the  following  typical  form,  which  is  called 
the  General  Equation  of  the  nth  Degree: 

cc«  +  Ax"-'  +  Bx"-^  +  Cx"-^  + +  A-^O.  (1) 

In  this  the  coefficients  are  r§al  and  rational,  and  the 
exponents  positive  integers ;  the  coefficients  may  be  inte- 
gral or  fractional,  positive  or  negative.  The  equation  is 
arranged  according  to  the  descending  powers  of  the  un- 
known number :  K  is  called  the  Absolute  Term,  because  it 
does  not  contain  x  ;  but  it  is  classed  among  the  coefficients, 
since  it  may  be  regarded  as  the  coefficient  of  x°.  If  any 
one  of  the  coefficients  is  zero,  the  corresponding  term 
vanishes. 

Reduce  to  the  typical  form, 

6a:*  -  2x'  -f  Sx^  +  20  —  14a;  =  37. 

The  coefficient  of  the  highest  power  of  x  must  be  made  unity. 
Transposing,  arranging,  and  dividing  by  —  2,  and  supplying  the 
missing  power  of  x  we  have, 

x^  —  Sx*-\-Ox^-  |a;2+  7x+  y- =  0. 

Here  ^  =  —  3,  B=0,  C  =  —  |,  D=7,  K  =  \s  and  n=5. 

An  Algebraic  Function  of  x  is  an  algebraic  expression 
whose  value  depends  upon  that  of  x.  A  function  of  x  is 
denoted  by  such  symbols  as  F(x),f(x),  and  <fi(x).     Thus, 

7x^  +  4.x^  +  2x-5 
is  a  function  of  x. 


458  ALGEBRA. 

We  shall  designate  equation  (1)  by  F(x)  =  0 ;  and  when 
its  coefficients  as  well  as  exioonents  are  all  integral,  we  shall 
denote  it  hy  f(x)  =  0. 

568.   A  Cubic  Equation  is  one  of  the  third  degree ;  as, 

x^  —  5x^-\-7  =  0. 

A  Biquadratic  Equation  is  one  of  the  fourth  degree ;  as, 
x'-3x'-{-4.x  —  15  =  0. 

A  Root  of  an  equation  is  an  expression,  which,  when 
substituted  for  the  unknown  number,  will  satisfy  the 
equation.  If  the  equation  is  in  the  typical  form,  the  first 
member  will  become  zero. 

Imaginary  roots  are  found  in  Higher  equations  as  well 
as  in  quadratics. 

If  an  equation  having  literal  coefficients  is  lower  than 
the  fifth  degree,  it  can  be  solved  by  methods  discovered  by 
Ampere,  Clausen,  Descartes,  Euler,  and  others.  Abel  has 
proved  that  such  an  equation  cannot  be  solved  if  its  degree 
is  higher  than  the  fourth. 

Equations  of  any  degree  can  be  solved  if  the  coefficients 
are  nuTnerical. 

669.  F{x)  is  exactly  divisible  by  x  —  a,  if  a  is  a  root  of 
F(x)=-0. 

For  the  division  can  be  continued  until  the  remainder 
does  not  contain  x.     Let  Q  represent  the  quotient,  and  B 
the  remainder.     Since  the  dividend  equals  the  product  of 
the  divisor  by  the  quotient,  added  to  the  remainder, 
(x  —  a)  Q-\-E  =  0. 

But,  by  supposition,  x  =  a;  hence,  (x  —  a)  Q  =  0,  and  so 
^  =  0.     Therefore,  the  division  is  exact. 

670.  Conversely,  if  x  —  a  is  an  exact  divisor  of  F(x),  a 
is  a  root  of  F(x)  =  0. 


EQUATIONS  IN  GENERAL.  459 

Let  Q  be  the  quotient,  then 

Q{x-a)=(). 

Now,  this  equation  is  satisfied  when  a  is  substituted  for 
X ;  hence,  by  definition,  a  is  a  root  of  the  equation. 

571.  If  a  is  a  root  of  F  (x)  =  0,  it  is  a  factor  of  the  abso- 
lute term. 

For,  if  a  is  a  root,  x  —  a  is  a  divisor  of  F(x)',  and,  by 
the  principles  of  division,  —  a  must  be  a  factor  of  the  abso- 
lute term ;  hence,  +  a  is  also  a  factor. 

672.    A  ready  method  of  determining  whether  a  given 
number  is  a  root  of  a  given  equation  is  furnished  by  §§  570, 
571.     Thus  it  may  be  shown  that  4  is  a  root  of 
a:*  — llic^  +  28a;8  +  2ar^  — 16ic  +  32  =  0. 

We  see  that  4  is  a  factor  of  the  absolute  term,  and  proceed  as 
follows : 

x6  -  1 1  x4  +  28  x3  +  2  x'-i  -  1 6  X  +  32  |x-4 

x5-    4x^ x4-7x8  +  2x-8 

-7x4  +  28x8 
-7x4  +  28x8 

+  2x2-16x 
+  2x-^-    8x 

—  8X  +  32 
-8X  +  32 

The  work  may  be  shortened  by  omitting  the  letters,  and 
using  the  numbers  only.  If  any  power  of  x  is  wanting,  it 
should  be  supplied,  the  coefiicient  being  zero.  The  work 
now  appears  as  below  : 

1-11+28  +  2-16  +  32  |l-4 

1—    4  1-7+2-8 


-7  +  28 

—  7  +  28 

+  2- 

-16 

+  2- 

-    8 
-8  +  32 
-8  +  32 

460  ALGEBRA. 

But  the  operation  may  be  still  further  abridged.  As  the 
first  term  of  the  divisor  is  unity,  the  first  term  of  each  re- 
mainder is  the  next  term  of  the  quotient ;  also,  the  product 
of  the  first  term  of  each  remainder  by  the  second  term  of 
the  divisor  gives  the  second  term  of  the  remainder.  The 
quotient  will  not  be  altered  if  we  use  1  +  4  as  a  divisor, 
and  substitute  addition  for  subtraction  in  the  foregoing 
process.     Thus  we  have,  omitting  the  first  coefiicient, 

-11  +  28  +  2-16  +  32 

4-28  +  0+    8  —  32 

-7    +0  +  2    -8    +0 

The  last  horizontal  line  contains  the  coefficients  of  x  in 
the  quotient,  except  that  of  the  highest  power,  which  is  1. 
The  number  0  is  the  remainder  which  comes  from  the 
division. 

If  the  coefficients  are  denoted  by  J,  B^  C,  etc.,  and  the 
trial  root  by  r,  we  have  the  following  rule  : 

W^'ite  A,  B,  C, ,  K  i7i  a  horizontal  line. 

Under  A  write  r,  and  write  their  sum  under  r. 

Multiply  the  sum  hy  r,  and  add  the  product  to  B  ;  multi- 
ply this  sum  hy  r,  and  add  the  product  to  C ;  and  so  on.  If 
the  last  sum  is  zero,  the  division  is  exact. 

This  method  is  called  Synthetic  Division. 

When  the  coefficient  of  the  first  term  of  the  dividend  is 
not  unity,  it  must  be  multiplied  by  r,  and  the  product 
added  to  A ;  this  sum  multiplied  by  r  is  then  added  to  B  j 
and  so  on. 

Exercise  133. 

Determine  whether  the  number  placed  in  a  parenthesis 
after  each  equation  is  a  root  of  the  equation : 

1.  x'-\-^x^  —  l()x^— 112x^  —  201  x-110  =  0.     (-5.) 

2.  x'-Sx'-\-lx^-{-x^—'6x-\-2  =  ^.     (1.) 


EQUATIONS  IN  GENERAL.  461 

3.  x'^21x  +  7x'  +  U7  =  0.     (-7.) 

4.  x'-{-Sx*r-7x'  —  54.x-\-16  =  0.     (-8.) 

5.  x'-4:x'  —  3x'  —  2x  —  S  =  0.     (2.) 

6.  x^-i-Ux'  +  65x  +  112  =  0.     (-7.) 

7.  2x'-4x'-62x'-\-lUx-lS0  =  0.     (6.) 

8.  x^-7x  — 2a;2  — 15  =  0.     (—5.) 

9.  x'-^2.Sx'-\-3.6x'-\-4:.9x-\-1.2  =  0.     (-0.3.) 

10.     ^3_,^.2_^x^^_l_0.       (f.) 

573.  If  F  (x)  =  (x  —  a)  (x  —  b)  (x  —  c),  etc.,  a,  b,  c,  etc., 
are  the  roots  of  F(x)^=0. 

For,  by  hypothesis, 

(X  —  a){x  —  b)  {x  —  c){x  —  d)....  =  0. 

This  equation  is  satisfied  if  x  =  a,  for  this  supposition  will  reduce 
the  first  member  to  zero.  The  equation  will  be  satisfied  if  x  =  a,  or 
if  X  =  6,  or  if  X  =  c,  or  if  x  =  d,  etc.;  so,  by  definition,  a,  6,  c,  d,  etc., 
are  roots  of  F  (x)  =  0. 

The  degree  of  the  equation  equals  the  number  of  bino- 
mial factors.  When  there  are  n  factors,  the  equation  is  of 
the  nth  degree ;  conversely,  when  the  equation  is  of  the 
nth  degree,  and  resolvable  into  factors  of  the  form  (x  —  a), 
there  are  n  of  those  factors. 

674.  When  F(x)=0  is  of  the  nth  degree,  it  has  n  roots 
and  no  more* 

The  equation  is,  §  567, 

x«  -h  Ax^-^  +  J5x«-2  +  Cx^-^  + +  ^=  0.  (1) 

*  Note.  It  is  assumed  that  every  equation  has  at  least  one  root. 
The  demonstration  of  this  fact,  which  is  long  and  difficult,  need  not 
be  given  here. 


462  ALGEBRA. 

Let  a  be  a  root  of  equation  (1);  then,  by  §  569,  the  equation  is 
divisible  by  x  —  a,  and  the  quotient  will  be  of  the  form 

x«-i  +  ^'x«-2  +  B'x^-^  +  C"x«-4  +  +  /r  =  0.  (2) 

Now  (2)  must  have  at  least  one  root,  and  let  b  be  that  root. 
Dividing  (2)  by  {x  —  b),  dividing  the  resulting  equation  by  (x  —  c),  c 
being  a  root  of  that  equation,  and  continuing  the  process  until  n  —  1 
divisions  have  been  made,  we  reach  a  result  of  the  form 

x-f=0.  (3) 

It  is  plain,  therefore,  that  the  first  member  of  equation 
(1)  is  the  product  of  n  factors  of  the  form  (x  —  a),  and  so, 
by  §  573,  it  has  n  roots.  If  there  were  more  than  n  roots 
there  would  be  more  than  n  factors,  and  the  degree  of  the 
equation  would  be  higher  than  the  nth.  These  roots  are 
not  necessarily  all  different.     The  roots  of 

x^  — cc^  — 8a;  +  12=:0 

are  2,  2,  and  —  3. 

This  equation  has  two  equal  roots. 

575.  When  the  roots  of  an  equation  are  given,  the  equa- 
tion is  readily  found. 

Find  the  equation  whose  roots  are  3,  7,  —  1,  and  —  2. 

(x-3)  (x-1)  {x^l)  (a:+2)=0; 

or,  by  multiplying, 

ic^— 7  a-^  —  7a^2_|_  43^.^42  =  0. 


Exercise  134. 
Find  the  equations  whose  roots  are  given  below : 

1.  2,  6,  and  —7.  3.    2,  3,  —2,-3,  and  —6. 

2.  1,  4,  —1,  and  —3.  4.    0.2,  \,  and  —0.4. 

5.    5,  3+ V^,  and  3- V^. 


EQUATIONS  IN  GENERAL.  463 

576.  The  relation  between  the  coefficients  and  the  roots 
of  an  equation  is  to  be  investigated. 

According  to  §  575,  a  and  h  are  the  roots  of 

x^  —  {a-\-h)x-\-ah  =  ^',  (1) 

a,  5,  and  c  are  the  roots  of 

x^  —  {(ji-\h^c)x'-\'{ab^ac^hc)x-abc  =  ^',  (2) 

a,  h,  c,  and  d  are  the  roots  of 

X* —  (a-\-h-\-c-\-d)  x^-\-  (ah  -{-  ac-\-  ad-\-hc-\-hd-]r  cd)  x^ 
—  (abc  +  ahd  +  acd  -\-  bed)  x  -\-  abed  =  0  ;     (3) 
and  so  on. 

From  these  the  following  conclusions  may  be  drawn: 

I.  The  coefficient  of  the  second  term,  with  its  sign 
changed,  equals  the  sum  of  the  roots ;  the  coefficient  of 
the  fourth  term,  with  its  sign  changed,  equals  the  sum 
of  the  different  products  that  can  be  formed  by  taking 
the  roots  in  groups  of  three ;  and  so  on. 

II.  The  coefficient  of  the  third  term  equals  the  sum  of 
the  different  products  that  can  be  formed  by  taking  the 
roots  in  groups  of  two  ;  the  coefficient  of  the  fifth  term 
equals  the  sum  of  the  different  products  that  can  be  formed 
by  taking  the  roots  in  groups  of  four ;  and  so  on. 

III.  The  absolute  term  equals  the  continued  product  of 
the  roots  with  their  signs  changed. 

Exercise  135. 

By  the  above  principles,  form  the  equations  whose  roots 
are  given  below : 

1.  2,  4,  and— 3.  4.    6,  6,  and  6. 

2.  2,  —1,  and  —7.  5.    2,  1,  —2,  and  —1. 

3.  2,  0,  and  —2.  6.    2,  i,  -  2,  and  —  i. 


464  ALGEBRA. 

Solutions  by  Factoring. 

577.   i'\x)  is  sometimes  factored  by  inspection,  and  the 
roots  of  F(x)=^0  are  thus  found: 

(1)  Solve  a;2  -  7  cc- 18  =  0. 

By  factoring,  we  have  {x  —  9)  (x  +  2)  =  0. 
Therefore,  by  §  570,  x  =  9,  or  -  2. 

(2)  Solve  ic^  — 8  =  0. 

(x-2)(x2  +  2x+'4)=0.  (1) 

Now,  as  equation  (1)  may  be  satisfied  by  putting  either  factor 
equal  to  zero,  we  have 

X  -  2  =  0,  (2) 

and  x2  +  2  X  +  4  =  0.  (3) 

From  (2)  we  see  that  x  =  2,  and  the  solution  of  (3)  gives 
X  =  -  1  +  V^,  or  —  1  -  V^. 

(3)  Solve4a;*  — a;2  4-2a;  — 1  =  0. 

4X4—  (X2-2X+  1)=:0, 

(2  x2)2  -  (X  -  1)2  =  0, 

(2  x2  +  X  -  1)  (2  x2  -  X  +  1)  =  0. 

2  x2  +  X  —  1  =  0.  (1) 

2  x2  -  X  +  1  =  0.  (2) 

From  (1),   X  =  i  or  —  1;  

from  (2)         X  =  i  (1  +  V-  7)  or  i  (1  -  V-  7). 


Exercise  136. 

Find  the  roots  of  the  following : 

1.  x^-\-llx-\-24.  =  0. 

2.  7  a;2  + 161  cc  +  714  =  0.     (Divide  by  7.) 

3.  a;^  — 4aV  +  3a*  =  0. 

4.  o;^  +  4  cc^  +  8  x2  +  32  =  0.     (Divide  by  x^  +  8.) 

5.  12x'-^x-2==0. 


EQUATIONS  IN  GENERAL.  465 

6.  4ic*  — 9ic-  +  6.T  — 1=0.  8.    a;«  — 64  =  0. 

7.  A9x^  —  112bx  +  6'ib^  =  0.  9.    3x"^  — a^2_^3a;  — 1  =  0. 

10.    a;  — 27a;*  =  0. 

578.  Cubic  equations  in  which  at  least  one  root  is  inte- 
gral are  easily  factored.  Let  — a  be  an  integral  root;  then 
the  equation  may  be  written  in  the  form 

(x  +  a)  (x^  +  mx  -f  n)  =  0,  (1) 

or          x^-\-(a-{-  m)  x^  +  (^^  +  «^^0  a?  -f-  a/i  =  0.  (2) 
a                      n 

7n                    am  am  -^  m  =  a. 

By  trial,  a,  m,  and  n  are  to  be  obtained.  Resolve  the  absolute 
term  into  two  trial  factors.  If  these  factors  are  the  correct  values  of 
a  and  m,  when  a  is  subtracted  from  the  coefficient  of  x-,  and  n  from 
the  coefficient  of  x,  as  above,  the  second  remainder  divided  by  the 
first  remainder  will  give  the  first  subtrahend,  as  shown  above.  In 
this  example,  a  is  the  first  subtrahend,  m  the  first  remainder,  and  n 
the  second  subtrahend  ;  —  a  is  one  root,  and  by  solving  x^+mx+n^O, 
the  other  roots  are  determined. 

Solve  cc^  —  9  x=^  +  26  a;  —  24  =  0. 

x3-l)x-  +  2{)X-24  =  0 
-3+8 

—  6+18  18  + -6  =  -3. 

.*.  X  —  3  =  0,  and  x-^  —  C  x  +  8  =  0  ;  whence,  x  =  3,  2,  and  4. 

Exercise  137. 
Solve  the  following : 

1.  a;3  +  3a:2  — 25x  — 12  =  0.         6.  a^^  — 3.t2  — 54x  — 104=0. 

2.  a^3  — 4a;2  — 8x  +  8  =  0.  7.  a;^  +  9x2  +  2a:-48=0. 

3.  ic^  — 7x-2+19a^  — 21  =  0.        8.  ic^  — 2a;2  — 25.:c-f50  =  0. 

4.  a;3  — 8a;2+21x  — 18  =  0.        9.  a:^  — 3a2  — 61a'  +  63  =  0, 

5.  a;3  — 26a;  — 5  =  0.  10.  a;^  — 37a;  — 84  =  0. 


466  ALGEBRA. 

679.  Biquadratic  equations  can  be  solved  if  they  can  be 
resolved  into  two  factors  of  the  second  degree,  even  when 
no  root  is  integral. 

(x^  -f"  '^^^  +  ^)  (^^  -\-px  -\-  q) 

=  x^-\-  (m  +i5)  ^^  +  (n  +  mp  -\-q)x^-^  (np  -\-  mq)  x  +  nq. 
Solve  a^*  +  13  a;3  +  33  a^2  +  31  a;  + 10  =  0. 
The  following  are  the  values  of  the  coefficients : 

m  +  i)=  13,       ,  (1) 

n+mp  +  g=  33,  (2) 

np  +  mq  =  31,  (3) 

and  nq  =  10.  (4) 

Take  trial-values  of  n  and  q  from  (4).  From  (2)  find  the  value  of 
mp.  Combine  this  with  (1),  and  get  by  inspection  the  values  of  m 
and  p.     If  the  values  of  m,  n,  p,  and  q,  satisfy  (3),  they  are  correct. 

First  Hypothesis. 

Let  n  =  —  2,  and  q=  —  6. 

Then,  from  (2),  mp  '=  40, 

and,  using  (1),  ?n  =  5,  and  p  =  8. 

These  values  will  not  satisfy  (3). 

Second  Hypothesis. 
Let  n  =  1,  and  q  =  10. 

Then,  from  (2),  mp  =  22, 

and,  using  (1),  m  =  2,  and  p=  11. 

These  values  satisfy  (3).  Since  the  two  factors  of  the  biquadratic 
are  {x^  +  nix  +  n)  and  {x^  +P^  +  Q')j  each  of  which  equals  zero,  we 
have,  in  this  example, 

x^+    2x+  -1  =  0, 
and  a:2  +  11 X  +  10  =  0; 

the  solution  of  which  gives  the  values  of  x. 

The  values  of  7i  and  q  can  be  interchanged  without  affect- 
ing the  values  of  m  and  p.  If  any  hypothesis  makes  m 
and  p  contain  surds,  it  is  to  be  rejected,  unless  n  =  q;  for, 
if  m  and  p  contain  surds,  (np-{-mq),  the  coefficient  of  x, 
will  contain  surds,  unless  n  =  q)  in  which  case  the  surds 
may  disappear. 


EQUATIONS    IN    GENERAL. 


467 


Exercise  138. 


Solve  the  following : 


1.  a; 

2.  X 

3.  X 

4.  X 

5.  X 

6.  X' 

7.  a; 

8.  X 

9.  ic 
10.  a; 


—  2x^  —  13x''-\-3Sx  —  24:  =  0. 

—  5x^  —  2x^-\-12x-{-8  =  0. 
-4:X^  —  8x-{-32  =  0. 

—  12x'-\-oOx'-S4.x-\-A9  =  0. 

—  llx-2+18a^  — 8  =  0. 

-10:i-"-20.r-16  =  0. 

-7x^  +  23x^  —  ^7x-]-4.2  =  0. 

+  2:i-^-9x2-8a:  +  20  =  0. 
-4x^  — 102;z;2_igg^,_91^() 

—  Ilic3  +  46a:2  — 117ic  +  45  =  0. 


Descartes'  Rule  of  Signs. 

680.  A  Complete  Equation  is  one  in  which  no  power  of 
X  is  wanting. 

If  two  successive  terms  have  like  signs,  there  is  a  2)erma- 
nence  of  sign  ;  if  they  have  unlike  signs,  there  is  a  variation. 
Thus,  in  the  complete  equation 

cc^  — a:«  — 5a;*  +  4ic*  +  2a;3  — 3ic2+6x  +  7  =  0, 

there  are  three  permanences  and  four  variations. 

Descartes  discovered  that  the  signs  of  the  roots  are 
related  to  the  signs  of  the  terms  of  an  equation. 

681.  Descartes'  Rule  (modified)  is  as  follows  : 

No  comjilete  equation  has  a  greater  number  of  positive 
roots  than  of  variations  of  sign,  nor  a  greater  number  of  neg- 
ative roots  than  of  permanences  of  sign. 


468  ALGEBRA. 

Suppose  that  the  successive  terms  of  a  certain  complete 
equation  have  the  signs 

+  -  +  +  +  --  + 

Here  are  three  permanences  and  four  variations.  To  in- 
troduce a  new  positive  root  into  the  equation,  we  multiply 
it  by  cc  —  %  The  resulting  signs  are  shown  in  the  follow- 
ing scheme : 

+  -  +  +  + + 

+  - 

+  -  +  +  + + 

-  + +  +  - 


-f-  +  ±±-±+- 

The  sign  zb  indicates  that  there  is  doubt  whether  the 
term  is  positive  or  negative.  Inspection  shows  that  where 
there  were  permanences  in  the  multiplicand,  there  are 
ambiguities  in  the  product.  Therefore,  the  introduction  of 
the  root  +  a  has  not  increased  the  number  of  permanences. 
But,  as  it  has  made  the  number  of  terms  one  greater,  the 
number  of  variations  must  have  been  increased  by  one,  at 
least. 

As  the  introduction  of  a  positive  root  increases  the 
number  of  variations,  the  whole  number  of  positive  roots 
cannot  exceed  the  number  of  variations.  The  root  — a  can 
be  introduced  into  an  equation  by  multiplying  lihj  x-\-a] 
and,  by  reasoning  similar  to  the  preceding,  it  can  be  shown 
that  the  number  of  negative  roots  cannot  exceed  the  num- 
ber of  permanences. 

682.  When  all  the  roots  of  a  complete  equation  are  real, 
the  numher  of  positive  roots  equals  the  number  of  variations, 
and  the  numher  of  negative  roots  equals  the  number  of  per- 
manences. 


EQUATIONS  IN  GENERAL.  469 

Let  jp  be  the  number  of  positive  roots,  n  the  number  of 
negative  roots,  v'  the  number  of  variations,  and  ]/  the  num- 
ber of  permanences. 

Now  the  degree  of  the  equation  equals  'p  +  n,  and  also 
equals  v^  -\-p\     Therefore, 

or,  by  transposing,      v' — ij=^n — p\  (1) 

Now,  by  §  581,  the  first  member  of  equation  (1)  must  be 
positive,  or  zero  ;  also  the  second  member  must  be  negative, 
or  zero.  Since  the  first  member  equals  the  second,  both 
are  zero.     Hence,  v'=p,  and  n^p^ 

683.  In  an  equation  which  lacks  one  power  of  cc,  the 
presence  of  imaginary  roots  may  sometimes  be  detected. 
For  illustration,  take  the  equation 

We  are  at  liberty  to  assume  that  the  second  term  is 
positive,  or  that  it  is  negative. 

Taking  it  as  positive,  tliere  is  no  variation,  and  the 
equation  has,  therefore,  no  positive  root. 

Assuming  the  second  term  to  be  negative,  there  is  only 
one  permanence,  and  so  there  cannot  be  more  than  one 
negative  root. 

As  there  are  three  roots,  and  as  imaginary  roots  come  in 
pairs  (§  586),  we  conclude  that  one  root  is  negative,  and 
two  are  imaginary. 

684.  A  complete  equation  whose  signs  are  all  positive 
can  have  no  positive  real  root,  for  there  are  no  variations 
of  sign.  When  the  signs  are  alternately  positive  and  nega- 
tive, there  are  no  negative  real  roots,  for  there  are  no 
permanences  of  sign. 


470  ALGEBRA. 

Exercise  139. 

All  the  roots  of  the  equations  given  below  are  real ; 
determine  their  signs. 

1.  x'-\-4.x^-4l3x'  —  5Sx-[-24:0  =  0. 

2.  x^  —  2^x^-\-155x  — 350  =  0. 

3.  x^-\-4.x^  —  35x''—78x  + 360  =  0, 

4.  x^-12x'-4.3x  —  30  =  0. 

5.  x'-3x'  —  5x^^l5x^  +  4.x  —  12  =  0. 

6.  x^  —  12x^-^47x  —  60  =  0. 

7.  x'  —  2x^—13x''-]-3Sx  —  24:  =  0. 

8.  cc^  —  a;^  — 187x^  —  359x2  +  1860^  +  360  =  0. 

9.  x«  — 10x^  +  19x^  +  110x3  — 536x2 +  800x  — 384  =  0. 
10.  x'^-10x«+22x^  +  32x^-131x3+50x2+108x-72=0. 

Fractional  and  Imaginary  Koots. 
585.    A  rational  fraction  is  never  a  root  of  f(x')  =  0. 

Suppose  that  -,  a  simple  fraction  in  its  lowest  terms,  is  a 
root  of 

x'^  +  Jx"-^  +  ^x"-2+(7x«-3+ -i-K=0.  (1) 

Substituting  -  for  x,  and  multiplying  by  t"~^,  we  obtain 
r 

-  +  Al"-'  +  Btl"-'  +  CtH"-'  + +  Kt^-'  =  0.      (2) 

0 

But  the  sum  of  a  fraction  and  a  number  of  integers  can- 
not equal  zero.  Therefore,  equation  (2)  is  false,  and  -  is 
not  a  root  of  equation  (1). 


EQUATIONS  IN  GENERAL.  471 

686.    Imaginary  roots  enter  equations  in  conjugate  pairs. 

Any  imaginary  root  may  be  represented  by  the  expres- 
sion a  ±01.  (See  §  538,  and  the  last  sentences  of  §§  541 
and  546.)     When  a  is  zero,  the  root  is  a  pure  imaginary. 

There  must  be  an  even  number  of  imaginary  roots,  or  the 
absolute  term,  which  is  the  product  of  all  the  roots  with 
their  signs  changed,  would  be  imaginary ;  hence  these  roots 
are  in  pairs. 

The  product  of  x — (a-{-bi)  by  ic  —  (a — hi)  is  real.  If 
the  roots  could  not  be  grouped  in  conjugate  pairs,  the  prod- 
uct (x  —  a)  (x  —  h)  (x  —  c) would  involve  imaginaries. 

587.  If  F{x)  =^0  is  of  an  odd  degree,  it  has  at  least  one 
real  root ;  if  of  an  even  degree,  all  the  roots  may  he  imag- 
inary. 

An  equation  of  an  odd  degree  has  an  odd  number  of 
roots ;  hence,  all  the  roots  cannot  be  imaginary.  §  586. 

When  the  equation  is  of  an  even  degree,  it  can  be  resolved 
into  an  even  number  of  factors,  as  follows  : 

{x  —  a){x  —  h){x  —  c){x  —  d){x  —  e){x—f) =  0.   (1) 

If      {x  —  a)  {x  —  l))-=x--\-  'px-\-  q, 

and       (x  —  c)(x  —  d)^=x'^-\-  jh^  +  2ij  ^^^  so  ^^j 

equation  (1)  becomes 

(x^+px  4-  q)  (x^  +  PiX  +  2'i)  (x^+p^x  +  q^) =  0. 

Putting  each  of  these  trinomial  factors  equal  to  zero,  and 
solving  the  resulting  equations, 

if  ir^^'  ^^^    1  *^^i>  and  so  on, 

all  the  roots  will  be  imaginary. 


472  ALGEBRA. 


Equal  Roots. 

588.  In  finding  the  equal  roots  of  an  equation,  expres- 
sions called  Derivatives  are  used,  which  play  a  very  impor- 
tant part  in  the  Differential  Calculus, 

Let  a;  be  a  variable  (§  467).  If  it  is  gradually  increasing 
or  diminishing,  Sx  will  increase  or  diminish  three  times  as 
fast;  7x,  seven  times  as  fast;  and  mx,  m  times  as  fast; 
m  is  the  ratio  of  the  rate  of  change  in  mx  to  the  rate 
of  change  in  x. 

Thus,  if  x^S,  7ic  =  21.  Suppose  that  x  increases  from 
3.0  to  3.1,  then  F(x)  will  increase  from  21.0  to  21.7  in  the 
same  time.     That  is, 

the  increase  of  F(x)  _0.7_7 
the  increase  of  x         0.1      1 

But  while  x  increases  at  a  uniform  rate,  F(x)  may  not 
increase  at  a  uniform  rate. 

Thus,  if  F(x)=5x'—x^-{-4:X  —  l,  when  x  =  l,  F(x)  =  7', 
when  x  =  2,  F(x)  =  43  ;  when  x  =  3,  F(x)  =  137. 

Hence,  to  find  the  ratio  between  the  increments  of  F^x) 
and  of  X  at  the  instant  x  begins  to  change  from  any  par- 
ticular value,  the  increment  of  x  must  approach  indefinitely 
to  zero.     In  general, 

The  derivative  of  F(x)  is  the  limit  of  the  ratio  of  the  incre- 
ment of  F(x)  to  that  of  X  as  the  increment  of  x  approaches 
indefinitely  to  zero. 

An  increment  may  be  positive  or  negative. 

The  derivative  of  F(x)  may  be  found  as  follows : 

I.  In  F(x)  substitute  x-\-h  for  x,  h  being  an  increment 
that  approaches  zero  as  a  limit.  Denote  the  result  by 
F{x-\-h).  Subtract  F{x)  from  F(x-\-h):  the  remainder 
is  the  increment  of  F(x). 


EQUATIONS  IN  GENERAL.  473 

II.  Divide  this  remainder,  lF(x  +  h)  —  F(x)'],  by  h  :  the 
quotient  is  the  ratio  of  the  increment  of  F(x)  to  that  of  x. 

III.  Any  term  of  this  quotient  that  contains  A  as  a 
factor,  will  have  zero  for  a  limit,  and  therefore  vanish. 

589.  Find  the  derivative  of  aa^-{-b. 

ByL,  F(x)  =  ax^-\-b, 

and  F(x  +  h)  =  ax^  +  2  axh  +  ah''  +  b, 

.-.  F(x  +  h)  —  F(x)  =  2 axh  +  ah\ 

^    ^^         F(x-\-h)-F(x)      ^        .      . 
By  II,,       — ^^ ^ ^^=2  ax -{-ah. 

By  III.,     2 ax  is  the  derivative  of  ax^-\-b. 

Find  the  derivative  of  x\ 

(x-i-hy=x--\-2xh-i-h^j 

By  III.,     2a:;  is  the  derivative  of  x\ 
Find  the  derivative  of  x^. 

h 

By  III.,     Sx^  is  tlie  derivative  of  a;^ 

In.  general,  the  derivative  of  x"  is  nx"~^. 

For  x"  =  x  taken  ?i  times  as  a  factor;  and  if  x  increases, 
the  increase  of  each  x  is  multiplied  by  the  continued  product 
of  all  the  others  ;  that  is,  by  a:-""'.  But  there  are  n  of  these 
increasing  quantities,  and  the  derivative  is  therefore  nx^~\ 

Hence,  the  derivative  of  a  simple  expressioii  is  found  by 
multiplying  the  expression  by  the  exponent  of  the  variable, 
and  diminishing  the  exponent  by  unity. 

590.  When  F(x)  contains  a  number  of  terms,  involving 
different  powers  of  x,  the  derivative  of  the  entire  polynomial 
is  equal  to  the  sum  of  the  derivatives  of  the  separate  terms. 


474  ALGEBRA. 

For,  if  I\  /",  /"',  /'^,  etc.,  represent  the  increments  of 
the  successive  terms  of  the  polynomial  (when  x-\-h  is  sub- 
stituted for  ic), 

h  "^ 

is  the  derivative  of  F(x)  (when  h  approaches  indefinitely 
to  zero).     Likewise, 

/'    p;    i^'    i^ 

h      h       h       h 

are  the  derivatives  of  the  different  terms.     It  is  plain  that 

/' 4- /"  +  /'" +  7'^  .  —  I'l^'i:^!!!  I  ^  1  

h  h        h        h         h 

Find  the  derivative  oi  4.x^-\-Sx^-{-bx-\-l  =  0, 
Let  F{x)  =  4x3  +  3x2  +  5x  +  7. 
By  substitution, 

F{x  +  h)-  F{x)  =  12hx^+  {12h^-\-6h)x-\-4h^  +  Sh2+6h. 

•'•  ^^^  "^  ^^~  ^-  =  12x2  +  (12  A  +  6) X  +  4/i2  +  3^  +  5. 
By  letting  A  approach  zero  as  a  limit,  and  denoting  the  derivative 
by  F'{x),  p,^^^  =  12  x2  +  6  X  +  5. 

But  12x2,  6x,  and  5,  are  respectively  the  derivatives  of  4  x^,  3x2, 
and  5  x.  By  considering  the  last  term  of  F{x)  as  the  coefficient  of  x'', 
its  derivative  would  be  7  X  Ox-^,  which  equals  0. 

The  derivative  of  a  given  expression  is  called  its  first 
derivative ;  the  derivative  of  the  first  derivative  is  called 
its  second  derivative,  etc.  The  second  derivative  is  denoted 
by  F'\x)',  the  third  by  F'''(x),  etc. 

From  what  has  preceded,  the  following  rule  for  finding 
the  first  derivative  of  F{x)  is  obtained : 

Mxdtiphj  each  term  by  the  exponent  of  x  in  it,  and  dimin- 
ish the  exponent  by  unity. 


EQUATIONS  IN  GENERAL.  475 

591.   Find  the  successive  derivations  of 

F'    (x)  =  6x'-\-15x'-Sx^  —  lox^-\-4x-\-7, 
F"  (x)  =  S0x'-{-60x'  —  24:X^  —  S0x-i-4., 
F'"(x)  =  120x^  +  lS0x^  —  ASx  —  30, 
i^'v  (a^)z=  360^2  + 360  a; -48, 
i^^   (a;)  =  720x4-360, 
F"^'  (a-)  =  720, 
F''''(x)  =  0. 

Exercise  140. 
Find  the  successive  derivatives  of  the  polynomials : 

1.  x^-{-2x-^3.  3.    x*-{-2x^  —  5x'-\-64:. 

2.  x'-3x^-\-7x-\-25.      4.    a^-}-x*—6a^-\-3a^  —  4.x-\-2T. 

5.    X*  —  3ax^-\-6bx^  —  dcx-\-mn, 

692.  The  derivative  of  F(x)  is  the  coefficient  of  the  first 
power  of  h  in  F(x-\-  h). 

For,  F{x  +  h)  is  composed  of  F{x)  and  terms  each  of  which  con- 
tains some  power  of  h. 

F{x  +  h)  —  F{x)  is  composed  of  those  terms  of  F{x  +  h)  of  which 
each  contains  some  power  of  h. 

— ^ —  is  composed  of  the  coefficients  of  the  first  powers 

of  h  in  F{x  +  h),  and  of  terms  containing  h ;  rejecting  the  terms  con- 
taining h  (III.,  §  588),  the  derivative  of  F{x)  is  the  coefficient  of  the 
first  power  of  h  in  F(x  +  h). 

Find  the  derivative  of  F(x)  =  ax^  +  hx'^  -\-cx-\-d. 

Fix  +  h)  =  ax^  +  3  ax^h  +  3  axh^  +  a¥ 

+  bx^  +  2bxh  +  bh^ 

+  CX  +  ch 

+  d. 
F{x  +  h)  -  F{x)  =  (3ax2  -\-2bx  +  c)  h -h  {S ax  +  b)  h^  +  ah^, 


476  ALGEBRA. 

.:  F\x)  =  3  ax2  +  2  6x  +  c. 

593.  Apply  these  principles  to  the  detection  of  the  equal 
roots  of  F{x)  =  0. 

This  equation  may  be  written  in  the  form 

{x-a){x-h){x-c)\x-d) =  0.  (1) 

Substituting  x  +  ^  f  or  x, 

(X -  a  +  h){x  -  6  +  A)(x -  c  +  ^)(x  -d+h) ==0.      (2) 

If  (2)  be  expanded,  and  powers  of  h  higher  than  the  first  be 
neglected,  the  h  in  the  first  parenthesis  will  be  multiplied  by 

(x  —  6)  (x  —  c)  (x  —  d) ; 

the  h  in  the  second  parenthesis  will  be  multiplied  by 

{x  —  a){x  —  c){x  —  d) ;  and  so  on. 

The  total  coefficient  of  h  will  be 

(x-6)(x-c)(x-d) ] 

+  {x-a){x-c){x-d) I  (3) 

+  (X  —  a)  (X  —  h){x  —  d) J 

Now  (3)  is  the  derivative  of  F(x),  by  §  592. 

When  (1)  has  two  equal  roots,  let  a^=h.  We  see  that 
X  —  a  is,  in  this  case,  the  H.  C.  F.  of  (1)  and  (3).  When  (1) 
has  three  equal  roots,  let  a=^h=^c',  then  (x  —  ay  is  the 
,  H.  C.  F.  of  (1)  and  (3).  In  general,  when  i^(cc)  =  0  has  m 
roots  equal  to  a,  and  p  roots  equal  to  b,  the  H.  C.  F.  of  F(x) 
and  F\x)  is  {x  —  ay'-^{x—hy-^. 

594.  Hence,  in  seeking  the  equal  roots  of  F(x)=^(),  if 
<^  (x)  represents  the  H.  C.  F.,  it  will  be  seen  that : 

The  roots  of  cf>  (x)=^0  will  each  occur  once  more  in  F(x)=0 
than  in  <f>  (x)  =  0. 

When  there  is  no  H.  C.  F.  there  are  no  equal  roots.  In 
some  cases,  <^  (ic)  =  0  is  of  so  high  a  degree  that  it  cannot 


EQUATIONS  IN  GENERAL.  477 

be  solved  conveniently.      If,  however,  it  has  equal  roots, 
they  may  be  found  as  above. 

Solve  x^-^4.x^  —  20x^  —  50 x^  + 175 a^« 

+ 118  ic2  —  588  cc  +  360  =  0.  (i) 

F'{x)  =  7  x6  4-  24  x5  -  100  x*  -  200  x^  +  525  x2  +  236  x  -  588,    (2) 
0(x)  =  x3-x2-8x+  12, 

and  the  equation  to  be  solved  is, 

x3  —  x2  —  8  X  +  12  =  0.  (3) 

Apply  §  578  to  this  case,  or  else  proceed  to  find  the  equal  roots  of 
(3).     The  first  derivative  is 

3x2- 2x- 8.  (4) 

The  H.  C.  F.  of  (3)  and  (4)  is  x  -  2.  Since  x  -  2  =  0,  x  =  2,  and 
the  root  2  occurs  twice  in  (3).  Dividing  (3)  by  (x  —  2)2  gives  x  +  3  =  0; 
hence,  —  3  is  the  other  root  of  (3).  As  the  root  2  occurs  twice  in  (3), 
and  —  3  once,  we  know  that  (1)  has  as  roots,  2,  2,  2,  —  3,  and  —  3. 

By  dividing  (1)  by  (x  —  2)3  (x  +  3)2,  the  quotient  is 

x2  +  4x-5  =  0. 

The  solution  of  this  equation  gives  x  =  1  or  —  5.  The  seven  roots 
of  (1)  are,  therefore,  1,  2,  2,  2,  —  3,  —  3,  and  —  5. 

Exercise  141. 
Find  all  the  roots  of  the  following : 

1.  x^  —  Sx^-}-13x  —  6  =  0.  3.    x*  —  6x^  —  Sx  —  3  =  0. 

2.  x^— 7a;2_^16ic-12  =  0.        4.    a;^  — 2x-2  — 15ic4-36  =  0. 

5.  x'  —  7x^-\-9x^-{-27x-M  =  0. 

6.  ic^  — 24.^2 4-64a;  — 48  =  0. 

7.  a;^  — 10a;^  +  24a;2-f  lOic  — 25  =  0. 

8.  ic«  — 11^^  +  19x3+115x2  — 200a;  — 500  =  0. 

9.  x«  — 2x^  +  3x3-7x2  +  8x-3  =  0. 
10.  x^  +  6x^  +  x2-24x  +  16  =  0. 


478  ALGEBRA. 


Transformation  of  Equations. 

595.  The  solution  of  equations  is  usually  facilitated  by- 
reducing  them  to  the  form  /(cc)=0  ;  and,  since  we  treat 
only  of  equations  whose  exponents  are  positive  integers,  we 
have  simply  to  make  the  coefficient  of  the  first  term  unity, 
and  the  succeeding  coefficients  integral.  If  the  exponents 
were  negative  or  fractional,  the  equation  could  still  be 
reduced  to  the  form  f(x)  =  0. 

The  coefficient  of  the  first  term  may  be  reduced  to  unity, 
without  altering  the  values  of  the  roots,  by  the  following 
rule: 

Divide  the  equation  by  the  coefficient  of  the  highest  power 
of  X. 

When  this  division  makes  any  of  the  coefficients  frac- 
tional, the  equation  is  to  be  transformed  into  another  whose 
coefficients  are  integers. 

Let  m  be  the  L.  C.  M.  of  the  denominators  of  the  frac- 
tions. 

Substitute  —  for  x  in  the  equation 
m 

r^n  _|_  j^«-i  _^  ^^«-2  _^  cx^-s  ^ +  ^=  0,  (1) 

and  clear  of  fractions  ;  this  gives 

^f-\-Amif-^-\-Bmhf-''-\-CmY~^+ +^m«=0.   (2) 

The  coefficients  of  (2)  must  be  integral ;  because  m  is  the 
L.  C.  M.  of  the  denominators  of  A,  B,  C,  etc.  The  roots  of 
(2)  divided  by  ni  are  the  roots  of  (1) ;  for,  by  hypothesis. 

m 

Hence,  to  remove  fractional  coefficients. 
Multiply  the  coefficient  of  ic""^  by  m,  that  of  a;"~^  by  w^, 
etc.j  the  absolute  term  being  multiplied  by  m". 


EQUATIONS  IN  GENERAL.  479 

A  little  study  will  sometimes  enable  us  to  find  a  number 
less  than  the  L.  C.  M.,  whose  successive  powers  will  clear 
the  equation  of  fractions. 

Reduce  Sx^—4:X^-\-ix  —  ^  =  0 

to  the  form  /(cc)  =  0. 

Divide  by  3,  3^  — ^x^+  ^x  —  i  =  0. 

The  L.  C.  M.  of  the  denominator  is  12  ;  and,  by  the  last  rule, 

3.3  _  16x2  —  24x  -  432  =  0, 

which  is  in  the  form  required. 


Exercise  142. 
Put  these  equations  in  the  form  f(x)=0  : 

1.  2x^+^x'—x-\-^=0.        3.  5x*— a;«— J/x'-V-^+l=0. 

2.  3x^+5x^—l^x—S=0.     4.  x'-\-ix'+ix^—ix^-\-x~3=0. 

5.  x*  —  2x^-\-ix  —  U  =  0.     (Supply  the  term  Ox^) 

Hereafter  we  shall  treat  only  of  equations  which  are  in  the 
form  f(x)  =  0. 

696.  The  signs  of  the  roots  of  f(x')=^0  are  changed  by 
changing  the  signs  of  the  alternate  terms,  beginning  with  the 
second. 

By  §  576,  I.  and  II.,  it  is  plain  that  changing  the  signs 
of  the  roots  changes  the  signs  of  the  alternate  terms,  begin- 
ning with  the  second,  and  does  not  change  the  signs  of  the 
other  terms.  Conversely,  if  the  signs  of  the  alternate  terms 
of  a  given  equation  are  changed,  beginning  with  the  second 
term,  an  equation  is  obtained  whose  roots  are  the  roots  of 
the  given  equation  with  their  signs  changed.  Missing 
powers  of  x  must  be  supplied  with  the  coefficient  0. 


480  ALGEBRA. 

597.  The  roots  of  f(x)  =0  are  multiplied  by  m,  hy  multi- 
plying the  second  term  by  m,  the  third  by  m^,  etc. 

In  removing  fractional  coefficients,  we  multiply  x"~'^  by 
m,  x^~^  by  m^,  etc. ;  and  it  has  been  shown,  in  §  595,  that 
the  roots  of  the  resulting  equation  are  m  times  the  roots 
of  the  original  equation. 

598.  To  obtain  an  equation  whose  roots  are  the  reciprocals 
of  those  of  f(x)  =  0,  2vrite  the  coefficients  in  reverse  order. 

In  the  given  equation 

cc"  +  Ax""-^  +  ^ir"-2  + -i-Ix^-j-Jx-j-  K=  0,  (1) 

substitute  -  for  x,  multiply  by  ?/",  and  change  the  order 
of  the  terms  ;  the  result  is 

Xyn  +  J^n-^  _|.  7y«-2  _|_ +  ^^2  _^  J^  ^  1  _  Q.  (2) 

Comparison  of  (1)  and  (2)  shows  that  the  order  of  the 
coefficients  has  been  reversed. 

By  dividing  by  K,  (2)  may  be  reduced  to  the  form 
/(x)  =  0. 

Thus,  the  equation  whose  roots  are  the  reciprocals  of  the  roots  of 
x6  —  5a;5  -  7  x4  +  6x3  +  13x'^  -  9x  +  25  =  0, 
is  25x6  — 9x5+ 13x4 +  0x3-7  X--2- ox  +    i  =  q. 

599.  In  solving  some  higher  numerical  equations,  it  is 
necessary  to  transform  f(x')  =  0  into  an  equation  whose 
roots  are  less  by  A. 

In  the  equation 

a;«  +  ^£c«-i  +  ^x«-2+ -j-j-x-j-ir=0,  (1) 

by  putting  x  =  y-{-h  (in  which  case  y  =  x  —  A,  as  required), 
(y-^hy+A(y-\-hy-'+B(y-\-hy-'i- -{-J(y^h)+K=0.   (2) 

Expanded  by  the  binomial  theorem,  equation  (2)  becomes 


y^  +  A 
-\-nh 


yn-l  _|_  j5 

-}-(n  —  l)Ah 
+  in(n  —  l)h^ 


+ +^=0.      (3) 


EQUATIONS  IN  GENERAL.  481 

Let  A  +  7ih  =  A',  B  +  (n  —  l)Ah-{-^u(n  —  l)h^  =  B', 
etc.,  and  x  —  A  =  2/  5  equation  (3)  then  reduces  to  the  form 

(x  —  hy-\-A'(x  —  hy-^ 
+  B'(x  —  A)«-2  + +  J'(x  —  h)  +  K'  =  0.  (4) 

Divide  the  first  member  of  (4)  by  (x  —  h),  and  the  re- 
mainder is  K' ;  divide  the  quotient  just  found  by  (x  —  h), 
and  the  remainder  is  J'.  By  continuing  the  divisions,  all 
the  coefficients  of  (4)  can  be  found. 

The  values  of  x  are  the  same  in  (4)  and  (1)  ;  therefore, 
the  first  member  of  (4),  when  developed,  equals  the  first 
member  of  (1).  Hence,  if  the  successive  divisions  were 
performed  upon  (1),  the  remainders  would  be  the  same. 

600.  The  rule,  therefore,  for  transforming  f(x)  ==  0  into 
an  equation  whose  roots  are  less  by  h,  is  as  follows  : 

Divide  f(x)  hij  x  —  A,  and  the  remainder  will  he  the  abso- 
lute term  of  the  transformed  equation.  Divide  the  quotient 
just  found  by  x  —  A,  and  the  remainder  will  be  the  coefficient 
of  the  last  term  but  one  of  the  transformed  equation.  Con- 
tinue the  process  until  all  the  coefficients  are  determined. 

By  using  x-\-h  as  the  divisor,  f(x)=0  can  be  transformed 
into  an  equation  whose  roots  are  greater  by  h. 

The  above  rule  will  apply  when  the  coefficient  of  x"  is  not 
unity.     It  is  best  to  employ  Synthetic  Division  (§  572). 

Transform  into  an  equation  whose  roots  are  less,  by  5, 
x*  —  4.x^  —  7x^-\-22x-}-2^  =  0. 

-4-    7+    22  +  24 

+  5  +    5  -    10  +  60 

+  1  —    2  +    12  +  84  (84  =  first  remainder). 

+  5  +  30+140 


+  6  +  28  +  152  .  .  .  (152  =  second  remainder). 
+  5  +  55 


+  11  +  83 (83  =  third  remainder). 

+    5 

+  16 (IG  =  fourth  remainder). 


482  ALGEBRA. 

The  required  equation  is,  therefore, 

2/*  +  16  2/3  +  83  2/2  +  152  2/  +  84  =  0. 

The  roots  of  this  equation  are  —  1,  —  2,  —  6,  and  —  7  ;  the  roots  of 
the  original  equation  are  4,  3,  —  1,  and  —  2. 

Exercise  143. 

Transform  each  equation  below  into  another  whose  roots 
are  less  by  the  number  placed  in  the  parenthesis  directly 
after  the  equation : 

1.  a;3-ll:r^  +  31a;-12  =  0.     (1.) 

2.  x'-6x'  +  4.x'  +  lSx-5  =  0.     (2.) 

3.  x'-\-10x'-\-13x-24.  =  0.     (-2.) 

4.  0-3-9x2 +  220^-12  =  0.     (3.) 

5.  x'-\-x'-16x'  —  4.x-^  4.8  =  0.     (4.) 

6.  x^  +  2x3-25a^2_26a;  +  120  =  0.     (0.7.) 

7.  x'  +  x^  —  3x  +  4.  =  0.     (0.3.) 

8.  x'i-x'-{-3x'-2x  —  16  =  0.     (0.5.) 

9.  x^-3x^-2a:«  +  3x2-7ct:  +  12  =  0.     (-1.) 
10.  x^  —  x'-^2x'  —  3x^-{-4:x''—5x-{-6  =  0.     (0.2.) 

601.    To  transform  f(x')  =  0  into  an  equation  whose  second 

term  is  wanting,  substitute  y jor  x. 

In  (3)j  §  599,  to  make  the  second  term  vanish,  nh  must 

equal  —  A,  and  so  h  must  equal In  that  case,  x  = 

A  ^ 

y Hence,   to  make   the   second   term  vanish,  sub- 

^  A 

stitute  y for  x. 

^      n 


CHAPTER  XXXiy. 

Higher  Numerical  Equations. 

situation  of  the  roots. 

602.  A  commensurable  root  is  rational,  and  is  either 
integral  or  fractional. 

An  incommensurable  root  is  a  real  root  which  is  not 
commensurable. 

In  the  solution  of  higher  numerical  equations,  we  first 
determine,  by  §  572,  the  commensurable  roots.  These  are 
integral,  when  the  equation  is  in  the  form/(ic)  =  0  (§  585). 
In  finding  the  first  root,  which  we  shall  denote  by  a,  we 
get  the  coefiicients  of  the  equation 

^  =  0; 

X  —  a 

we  then  find  a  commensurable  root,  J,  of  that  equation,  and 
get  the  equation 

/(-) =0. 

(x  —  a)(x  —  h) 

Each  division  depresses  the  degree  of  the  equation  divided, 
and  when  all  the  commensurable  roots  have  thus  been 
divided  out,  if  the  resulting  equation  is  of  too  high  a  degree 
to  be  solved  easily,  we  proceed  to  find  the  incommensura- 
ble roots  by  Horner's  Method.  Imaginary  roots  may  then 
be  sought ;  the  student,  however,  will  not  be  required  to 
search  for  these. 


484  ALGEBRA. 

603.   Find  the  commensurable  roots  of 

x^-\-3x'  —  2x*  —  15x^  —  15x'-\-8x-{-20  =  0. 

The  factors  of  20  are  =h  1,  ±  2,  d=  4,  ±  5,  and  ±  10.  We 
try  1  at  first,  as  follows  : 

+  3-2-15-15+   8  +  20 
+  1  +  4+   2  —  13-28-20 

+  4  +  2  —  13-28-20        0 

Hence,  1  is  a  root ;  and  the  last  line  contains  the  coeffi- 
cients (except  the  first)  of  the  equation  obtained  by  divid- 
ing the  original  equation  by  x  —  1.     Next  try  —  1. 

+  4  +  2  —  13  —  28  —  20 
-1-3+  1+12  +  16 
+  3  —  1-12  —  16—   4 

Hence,  —  1  is  not  a  root.  A  trial  of  2  fails.  We  then 
try  —  2.  _}_  4  _j_  2  _  13  —  28  —  20 

-2  —  4+  4  +  18  +^ 
+  2-2-   9-10        0 

Hence,  —  2  is  a  root ;  and  the  last  line  contains  the  co- 
efficients of  the  next  depressed  equation.  A  trial  of  2  fails, 
but  —  2  succeeds,  giving  as  the  coefficients  of  the  depressed 
equation  q  2  5 

Trials  of  the  factors  of  —  5  all  fail.  The  roots  so  far 
found  are  1,  —  2,  and  —  2. 

The  equal  roots  might  have  been  found  by  §  594,  but  the 
work  would  have  been  tedious.  Three  exact  divisions  have 
been  made,  depressing  the  degree  of  the  original  equation 
by  three,  and  the  last  equation  is,  therefore, 

x^-i-0x^  —  2x-5  =  0, 

whose  roots  are  incommensurable. 


higher  numerical  equations.  485 

Exercise  144. 

Find  the  commensurable  roots  of  each  equation  below. 
The  number  of  these  roots  is  given  in  the  parenthesis. 

1.  x'  —  ix^  —  Sx-\-32  =  0.     (2.) 

2.  x^  —  Gx-  +  10x  —  S  =  0.     (1.) 

3.  a^-{-2x^  —  7x^  —  Sx-\-12  =  0.     (4.) 

4.  a:*^4-3a;2  — 30x  +  36  =  0.     (1.) 

5.  a-^-12.r^  +  32a:=^  +  27a--18  =  0.     (2.) 

6.  x'  — dx''-^  17x^  +  27 x  —  (JO^O.     (2.) 

7.  x'-rjx'-\-3x^-{-17x^-2Sx-\-12  =  0.     (5.) 

8.  x'  —  10x^-^35x-  —  50x  +  24:  =  0.     (4.) 

9.  x'  —  Hx*+nx^-\-29x^  —  36x  —  4o  =  0.     (3.) 
10.  x'  —  x^  —  6x^-\-9x--\-x  —  4:  =  0.     (1.) 

604.  After  the  commensurable  roots  are  removed,  the 
situation  of  the  incommensurable  roots  is  to  be  determined. 
Many  methods  of  doing  this  have  been  discovered,  the  most 
noted  of  which  is  Sturm's  Theorem ;  but  the  application  is 
tedious,  and  it  is  generally  best  to  proceed  by  trial. 

605.  If  two  trial  values  of  x,  substituted  in  /(x),  give 
results  with  tinlike  signs,  at  least  one  root  lies  between  the 
values.  If  the  sujiis  of  the  results  are  alike,  no  root,  or  some 
even  number  of  roots,  lies  between  the  values. 

Let  the  trial-values  of  x  be  m  and  n,  and  the  resulting 
values  oi  f{x)  be  J/ and  N,  which  have  unlike  signs.  If  x 
be  supposed  to  vary  continuously  from  ni  to  n,  f(x)  will 
pass  from  M  to  N,  and  must  pass  through  zero,  because  M 
and  N  have  unlike  signs. 

The  value  of  x  which  makes  f(x)  ^  0  is  a  root ;  hence,  at 


486  ALGEBRA. 

least  one  root  lies  between  m  and  n.  f(x)  may  pass  through 
zero  any  odd  number  of  times ;  but  if  the  number  of  times 
is  even,  M  and  N  will  have  like  signs. 

Let  M  and  N  have  like  signs.  When  x  varies  from  m 
to  71,  f{x)  may  not  pass  through  zero  at  all.  It  cannot  pass 
through  it  an  odd  number  of  times,  for  M  and  N  would 
then  have  unlike  signs.  It  may  pass  any  even  number  of 
times ;  and  therefore  no  root,  or  an  even  number  of  roots, 
lies  between  m  and  n. 

These  theorems  are  beautifully,  illustrated  by  the  loci  of 
equations.  Thus,  if  any  two  abscissas  are  chosen  (repre- 
senting values  of  x),  the  signs  of  the  corresponding  ordi- 
nates,  which  represent  values  of  /(x),  are  unlike  when  the 
number  of  intersections  of  the  curve  with  the  axis  of  X  is 
odd;  the  ordinates  have  like  signs  when  the  number  of 
intersections  is  even. 

606.  The  value  of  f(x),  ivhen  x  =  m,is  the  remainder  left 
after  dividing  f(x)  by  (x  —  ??i). 

Let  Q  be  the  quotient,  and  B  the  remainder  which  does 
not  contain  x.     Then 

f(x)  =  {x-m)Q-\-B. 
Now,  when  m  is  substituted  for  x, 

f(x)ovf(m)  =  K 
Thus,  when  x  =  5,  the  value  of 

x'  —  4.x^-^7x^  —  2x-}-AT 
is  computed  by  §  572  as  follows : 

—  4  +  0+  T—  2+  47 
+  5  +  5  +  25  +  160  +  790 
+  1  +  5  +  32  +  158  +  837 

Therefore, /(a:;)  =  837  when  x  =  5,  and  this  value  is 
obtained  more  expeditiously  than  by  direct  substitution. 


higher  numerical  equations.  487 

Exercise  145. 

Compute  the  value  of  each  expression  below,  the  value 
of  X  being  enclosed  in  the  parenthesis  following  each : 

1.  x^-hx^^2^x^  —  ^x^l.     (5.) 

2.  x^- 4x2  +  5^-22.     (—7.) 

3.  x'>-2x^^Zx^-\-x'-2^.     (2.) 

4.  a;^  +  7x«  — 2x2  — 49.     (_3>^ 

5.  a;^— 14a,'^  +  473.     (6.) 

6.  a;^  — 2a-^+3a^*  +  2x«  +  a;2  — 7a;-96.     (—2.) 

7.  a;"  — x^-2a:^  +  x«  — 6a^  +  14.      (3.) 

8.  x-^-4.T^  +  2a;2  — 7a;  +  16.     (10.) 

9.  x'  —  x^  —  2x^  —  Zx^\2x^-\-x?  —  x^\.     (-2.) 
10.  ic'  — 5a:*+6a;3  +  3.z;  — 1.     (4.) 

607.  Let  it  be  required  to  find  the  situation  of  the  roots 
of 

x*  — 2x3  — llx-  +  6a:  +  2==0. 

They  are  real.  Since  there  are  two  variations  and  two 
permanences  of  sign,  there  are  two  positive  and  two  nega- 
tive roots.  Inspection  of  the  signs  and  coefficients  shows 
that  these  roots  are  small  numbers. 

The  values  of /(x)  may  be  computed  by  §  606,  assuming 
values  of  x  from  — 4  onward,  until  the  situations  of  all  the 
roots  are  discovered.     These  values  are  tabulated  below : 


If  X  is 

/(x)  is 

If  xis 

f{x)  is 

-4 

+  186 

+  1 

-4 

-3 

+  20 

+  2 

-30 

2 

-22 

+  3 

-52 

—  1 

-12 

+  4 

-22 

0 

+  2 

+  5 

+  132 

488  ALGEBRA. 

The  changes  of  sign  in  the  columns  headed  f(x)  show 
that  the  lirst  root  lies  between  —  3  and  —  2,  the  second 
between  —1  and  0,  the  third  between  0  and  + 1,  and  the 
fourth  between  +  4  and  +  5.  When  any  root  lies  between 
0  and  ±1,  before  using  Horner's  Method,  it  is  necessary 
to  find  the  first  significant  figure. 

Search  for  the  root  between  0  and  -f  1 : 

If  a;  =  0.5,  f(x)=-^2.1. 

"  ic  =  0.6,  f(x)=-\-l.^. 

ii  x  =  0.1,  /(.x')=  +  0.4. 

"  a:  =  0.8,  /(ic)=-0.9. 

The  root  therefore  lies  between  0.7  and  0.8. 
The  root  between  0  and  —  1  is  next  sought : 

If  a;  =  -0.2,    /(ic)  =  +  0.4. 
"  x  =  —  0.^,    f(x)=  —  O.T. 

This  root  therefore  lies  between  —  0.2  and  —  0.3. 
The  first  significant  figures  of  the  other  roots  are  evi- 
dently —  2  and  +  4. 

608.  In  the  preceding  example,  the  changes  of  sign  of 
/(ic)  led  to  the  easy  discovery  of  the  situation  of  the  roots. 
But  two  successive  values  of  f(x),  having  the  same  sign, 
may,  by  §  605,  have  two  or  some  other  even  number  of 
roots  between  them. 

Take  the  equation 

ic«  +  5a^2_7^_|_2  =  o, 

whose  roots  are  all  real.  The  law  of  signs  gives  two  posi- 
tive and  one  negative  root.  It  is  easy  to  see  that  the 
positive  values  of  x  do  not  exceed  +  2,  and  that  the  nega- 
tive value  is  about  —  5.     We  therefore  tabulate  values  of 


HIGHER   NUMERICAL    EQUATIONS.  489 

K  X  is       f{x)  is  If  x  is       f{x)  is 


+  2 

+  16 

-3 

+  41 

+  1 

+  1 

-4 

+  46 

0 

+  2 

-5 

+  37 

-1 

+  13 

-6 

+  8 

—  2 

+  28 

—  7 

-47 

One  root  is  —  6  +  ;  the  values  of  f(x)  indicate  by  their 
rapid  diminution  and  close  approach  to  zero  in  the  first 
part  of  the  table,  that  f(x)  becomes  zero  when  x  is  between 
Oand+1. 

lfx  =  OA,  f(x)=-^OX 
"  x  =  0.5,  f(x)  =  -0.1. 
«  a;  =  0.6,  f(x)=  —  0.2. 
«  a;  =  0.7,  flx)=- 0.1. 
"  cc  =  0.8,    /(x)  =  +  0.1. 

The  other  roots  are  therefore  0.4+  and  0.7 +  . 

609.  The  figure  sought  may  also  be  found  by  a  method 
which  depends  upon  the  principles  of  differences.  Take 
the  example  in  §  607.  In  the  table  below,  the  column 
headed  di  contains  the  differences  of  the  first  order ;  that 
headed  ^4  contains  the  differences  of  the  second  order ; 
and  so  on.  The  differences  which  are  of  the  same  order  as 
the  degree  of  the  equation,  are  constant,  and  furnish  a  test 
of  the  accuracy  of  the  computation  of  the  values  oif(x). 

d\  di  dz  d^ 

+  10  +f  -48 

+  14  +''  _^4  +24 

-C  -20  0  +24 

-26  -f  +24  +24 


X 

/{^) 

-3 

+  20 

-2 

—  22 

-1 

-12 

0 

+  2 

+  1 

-4 

+  2 

-30 

+  3 

-52 

-22 


+  4 


490  ALGEBRA. 

In  Example  (3),  §  507,  w  =  f,  and  it  is  required  to  find 
the  number  which  added  to  6798.9  will  give  the  answer. 
If  the  answer  8331.4  had  been  given  and  n  required,  the 
following  equation  might  have  been  written : 

8331.4 =6798.9+^ (2029.7)  +  "^^""""^^  (123.1)  + ' 

or,  in  a  general  form, 

,   n(n  —  l) 
A  =  a-\-na^-\ ^—^ '-  a^  + j 

the  solution  of  which  would  give  an  approximate  value 
of  n. 

To  find  the  first  figure  of  the  root  which  lies  between 
0  and  -[- 1  in  the  present  example, 

A  =  f{x)^^',  a  =  +  2;  «!==  — 6;  a2  =  -20. 
The  equation  is 

0  =  2-6«-20^^^- ; 

or,  reducing, 

10^2  — 4w  =  2; 

whence  ?i=:0.7,  roughly;  and,  adding  this  to  0  (the  value 
of  X  which  stands  opposite  a),  0.7  is  an  approximate  value 
of  one  of  the  roots. 

The  next  figure  of  the  root  which  lies  between  —  3  and 
—  2  may  be  obtained  by  means  of  the  equation, 

0  =  20-427^  +  52 ''^\~^^; 
or,  reducing, 

267^'  — 68w=r=  — 20; 

whence,  71  =  0.3,  roughly;  and,  adding  this  to  — 3  (the 
value  of  X  which  stands  opposite  a),  —  2.7  is  an  approxi- 
mate value  of  this  root. 


HIGHER    NUMERICAL    EQUATIONS.  491 

610.  By  applying  principles  already  learned,  and  by  the 
exercise  of  due  discretion,  the  student  will  find  little  diffi- 
culty with  the  following  examples.  Skill  is  acquired  only 
by  practice. 

Exercise  146. 

Determine  the  first  significant  figure  of  each  root  in  the 
following  equations  : 

1.  a^  —  x^  —  2x-{-l  =  0,  6.  x^  —  6x'-\-3x-\-5  =  0. 

2.  x^  —  5x  —  3  =  0.  7.  .'c3  +  9A-2  +  24ic-f  17  =  0. 

3.  x^-5x'-{-7  =  0.  8.  x^-15x^-^63x-50  =  0. 

4.  ic^  — 7ic  +  7=:0.  9.  x'—Sx^+Ux^-{-4x—S=0. 

5.  0^3+2x2— 30a;  +  39=0.  10.  x*  — 12a;2-f  12a:  — 3=0. 

611.  In  difficult  cases,  where  the  roots  are  widely  differ- 
ent in  value,  Sturm's  Theorem  is  sometimes  useful ;  and  it 
is  given,  without  proof,  simply  for  reference : 

Divide  out  the  equal  roots  of  the  equation,  and  let  the 
resulting  equation  be  f(x)=0.  Let  fi(x)  be  the  first  de- 
rivative of  f(x).  Perform  the  operations  for  finding  the 
H.  C.  F.  of  fi(x)  and  f(x),  changing,  however,  the  sign  of 
each  remainder  before  using  it  as  a  divisor.     Denote  these 

successive  remainders  thus  changed  by /2(£c), /3(ic), ,  and 

let  r  be  the  final  remainder  with  its  sign  changed;  this 
remainder  does  not  contain  x. 

/(ic), /i(a:), /2(ir), ,  are  called  Sturm's  Functions. 

Let  a  and  h  be  trial-values  of  x.  Assuming  that  x=^a, 
compute  in  order  the  value  of  Sturm's  Functions,  and  note 
the  number  of  variations  of  sign  in  these  values  ;  likewise, 
note  the  number  of  variations  when  x  =  b.  The  difference 
between  the  number  of  variations  in  the  first  case  and  the 
number  in  the  second  equals  the  number  of  real  roots 
"between  a  and  h. 


492  ALGEBRA. 


Horner's  Method. 

612.  Having  found  one  or  more  figures  of  the  root,  we 
employ  for  the  remainder  of  the  work  Horner's  Method, 
which  was  discovered  in  the  early  part  of  this  century  by 
W.  G.  Horner,  of  Bath,  England.  This  process  will  be 
explained  by  the  solution  of  an  example. 

Take  the  equation 

/^'"  x'  —  (jx'-{-3x-}-5  =  0.  (1) 

Suppose  that  we  have  found,  by  one  of  the  methods 
already  given,  that  a  root  of  this  equation  lies  between  1 
and  2.  By  the  method  of  §  600,  we  transform  the  equa- 
tion into  one  whose  roots  are  less  by  1. 

-6  +3  +5 
+1  -5  -2 
^5        —2        +3. 

±1      zi 

—  4         -6 
+  1 

—  3 

The  transformed  equation  is,  therefore, 

yS_Sf-6y  +  3  =  0;  (2) 

or,  ey  =  3-3i/-i-f. 

Neglecting  for  the  moment  the  terms  — Sy^-{-y^,  since 
(as  will  be  seen  presently)  their  numerical  value  is  not 
large,  we  find  y^O.S.  With  this  assumed  value  of  y,  com- 
puting the  value  of  —3y^-\-y^,  and  substituting,  we  obtain 

6y  =  2.375', 

whence   y  =  OA,  approximately.     We  next  transform  (2) 
into  an  equation  whose  roots  are  0.4  less. 


HIGHER    NUMERICAL    EQUATIONS.  493 


—  3 

-6 

+  3 

+  0.4 

-1.04 

-2.816 

-2.6 

-7.04 

+  0.184 

+  0.4 

-0.88 

-2.2 

—  7.92 

+  0.4 

—  1.8 

The  second  transformed  equation,  whose  roots  are  0.4 
less  than  those  of  (2),  and  1.4  less  than  those  of  (1),  is, 
therefore, 

z^  —  l.Sz^  —  7.92z-\-0.1S4:  =  0.  (3) 

Neglecting  the  first  two  terms,  because  ^  is  a  small  deci- 
mal, and  z^ — 1.8;^-  is  still  smaller,  we  find  ^  =  0.02  +  . 

Equation  (3)  is  now  to  be  transformed  into  an  equation 
whose  roots  are  0.02  less. 


-1.8 

-7.92 

+  0.184 

+  0.02 

-0.0356 

-0.159112 

-1.78 

-7.9556 

+  0.024888 

+  0.02 

-0.0352 

-1.76 

-7.9908 

+  0.02 

-1.74 

The  third  transformed  equation,  whose  roots  are  1.42  less 
than  those  of  (1),  is,  therefore, 

r«-1.74t;2-7.9908t;  + 0.024888  =  0;  (1) 

whence  v  =  0.003  +  .  As  the  root  of  (4)  is  1.42  less  than 
the  root  of  (1),  the  root  of  (1)  must  be  1.42  + 0.003 +  ,  or 
1.423 +  .  The  preceding  process  can  be  continued  until 
the  root  of  (1)  is  found  to  any  required  degree  of  accuracy. 
Since  the  values  of  y,  z,  v,  etc.,  are  to  be  added  to  the 
first  rough  value  of  x,  they  are  all  positive  ;  and  hence  the 


494  ALGEBRA. 

coefficient  of  the  first  power  of  the  unknown  quantity  in 
each  transformed  equation  is  unlike  in  sign  to  the  absolute 
term.  If,  in  (4),  the  signs  of  the  last  two  terms  were  alike, 
the  value  of  v  would  be  —  0.Q03  +  .  This  would  show  that 
the  value  assumed  for  z  was  too  great,  and  we  should 
diminish  the  value  of  z  and  make  the  last  transformation 
again.  In  beginning  an  example,  one  is  very  likely  to 
assume  too  large  a  value  for  the  next  figure  of  the  root ;  in 
solving  (2),  for  instance,  the  first  solution  gave  ?/  =  0.5, 
and,  had  that  value  been  tried,  it  would  have  proved  to  be 
too  great. 

613.  It  is  not  necessary  to  write  out  the  transformed 
equations  in  full,  as  in  the  last  section.  But  when  the  co- 
efficients have  been  computed,  the  next  figure  of  the  root 
may  be  found  by  dividing  the  last  coefficient  (the  absolute 
term),  with  its  sign  changed,  by  the  coefficient  which  pre- 
cedes it.  Thus,  instead  of  writing  out  (4)  §  612,  we  might 
have  obtained  the  value  of  v  by  simply  dividing  — 0.0248  + 
by  —7.99  +  .  Indeed,  the  following  will  be  found  to  be  a 
good  practical  rule  : 

Whenever  a  true  significant  decimal  figure  of  the  root 
has  been  found  by  trial,  and  the  figure  found  by  dividing 
the  last  coefficient  by  the  coefficient  which  precedes  it 
proves  to  be  the  same,  the  subsequent  figures  of  the  root 
may  also  be  successively  found  by  dividing  the  last  coeffi- 
cients of  the  succeeding  transformed  equations  by  the 
coefficients  which  precede  them. 

If  n  decimal  places  are  required  in  the  root,  the  num- 
bers in  the  last  column  of  the  following  scheme  need  not 
be  carried  further  than  the  nth  decimal  place.  In  the 
other  columns,  a  sufficient  number  of  decimals  must  be 
employed  to  insure  the  accuracy  of  the  nth  place  of  the 
la§t  column, 


HIGHER   KUMERICAL   EQUATIONS. 


495 


-6                               +3                               +5 

1.423  + 

1                              —5 
-5                                -2 

—  2 
3 

1 
—  4 

-4 
-6 

-2.816 
0.184 

1 
-3 

-1.04 
-7.04 

-0.159112               1 

U.4: 

-2.6 

—  u.oo 
-7.92 

-  0.023988 
n  nnnoAn 

0  4 

-2.2 
0.4 

-1.8 

-  7.9556 

-  0.0352 

-  7.9908 

0.02 
-1.78 

0.02 
-1.76 

-  0.0052 

-  7.9960 

-  0.0052 

-  8.0012 

0.02 

-1.74 

0.003 
-1.737 

0.003 
-1.734 

0.003 
-1.731 

The  coefficients  of  each  transformed  equation  are  readily 
discerned  by  observing  the  lines  which  separate  the  differ- 
ent parts  of  the  work. 

The  next  three  figures  of  the  root  can  now  be  found  by 
dividing  —  0.000900  by  —  8.0012,  giving  0.000112  + .  The 
reason  for  this  is  shown  by  consideration  of  the  last  trans- 
formed equation, 

w^  — 1.731  w^  —  8.0012  w  +  0.000900  =  0. 

As  the  value  of  w  is  about  0.0001,  the  value  of  ^  — 
1.731  ?/;2  is  only  about  —0.00000002;  therefore  we  may 
assume,  without  sensible  error,  that 

8.0012  «/;  =  0.000900: 


whence 


=r0.000112. 


as  above. 


496  ALGEBRA. 

614.  One  of  the  roots  of  the  equation  of  §  612  is  —  0.6  +  ; 
and  to  avoid,  as  far  as  possible,  the  inconvenience  of  com- 
puting with  negative  numbers,  it  is  customary  to  transform 
the  equation  into  one  whose  root  is  +0-6  +  -  ^J  §  596, 
the  transformed  equation  is 

a  root  of  which  is  0.66966+  ;  hence,  a  root  of  the  original 
equation  is  —  0.66966 +  . 

615.  Sometimes  the  coefficient  of  the  first  power  of  the 
unknown  number  in  one  of  the  transformed  equations  is 
zero.  To  find  the  next  figure  of  the  root  in  this  case, 
divide  the  absolute  term  by  the  coefficient  of  the  second 
power  of  the  unknown  number,  and  extract  the  square 
root  of  the  quotient. 

For  instance, 

y^  — 7.6  ?/«  + 3.64  7/2  4-0  v/  — 0.0007829  =  0. 
whence,  3.64v/^  =  0.0007829,  approximately, 

and  y=0.01  +  . 

616.  In  rare  cases  two  of  the  roots  are  so  nearly  equal 
that  Horner's  Method,  carried  out  as  above,  will  not  find 
them  both.     Take,  for  example, 

^^  +  11:^=^- 102  a; +  181  =  0, 

in  which  we  have  found  that  there  are  two  roots  between 
3  and  4.  Horner's  Method  gives  for  the  first  transformed 
equation  f-^20f-92j-i-l  =  0. 

Since  the  coefficient  of  y^  is  large,  it  is  best  to  neglect  y^ 
only;  and  the  solution  gives  2/ =  0.2. 

The  next  transformed  equation  is 

z^  +  20.6z^  —  0.SSz-\-0.00S  =  0, 
and  —  0.008  -j-  -  0.88  =  0.009  + . 


HIGHER    NUMERICAL    EQUATIONS.  497 

Perceiving  that  20.6  z^,  when  transposed,  will  increase 
the  right-hand  member  numerically,  we  conclude  to  assume 
0.01  as  the  next  portion  of  the  root.  Proceeding  in  the 
usual  manner,  we  find  one  value  of  x  to  be  3.21312 +  .  The 
first  two  decimals  of  the  other  value  of  x  are  found  by  trial. 

Ific  =  3.21,    /(a:)  =  +  0.001261. 

"  a;  =  3.22,    f(x)  =  -  0.001352. 

''  x  =  3.23,    f(x)  =  -h  0.000167. 

The  change  of  sign,  and  the  numerical  values  of  f(x), 
show  that  the  second  value  of  a^  is  nearly  3.229. 

The  application  of  the  usual  process  gives  ic^  3.22952  +  . 

617.  When  the  root  sought  is  a  large  number,  in  finding 
the  successive  figures  of  its  interfral  portion  we  do  not 
divide  the  absolute  term  by  the  coefficient  of  the  first 
power  of  the  unknown  number,  because  the  neglect  of 
the  higher  powers,  which  are  in  this  case  large  numbers, 
would  lead  to  serious  error. 

Let  it  be  required  to  find  one  root  of 

a;^  -  3  a;2+ 11  a: -4842624131  =  0.  (1) 

By  trial,  we  find  that  a  root  lies  between  200  and  300. 
Diminishing  the  roots  of  (1)  by  200,  we  have 

if  +  800  if  +  239997  f  +  31998811  y  -  3242741931  =  0.  (2) 
liy  =  60,    /(//)  =  -  273064071. 
''  y  =  70,    /(y)  =  + 471570139. 

The  signs  of  /(^)  show  that  a  root  lies  between  60  and 
70.     Diminishing  the  roots  of  (2)  by  60,  we  obtain 

;s'»+1040;^^+405597.--+70302451.^— 273064071=0.   (3) 

The  root  of  this  is  found  by  trial  to  lie  between  3  and  4. 
Diminishing  the  roots  by  3,  we  may  find  the  remaining 
figures  of  the  root  of  the  original  equation,  which  are 
decimal,  by  the  usual  process. 


498 


ALGEBRA. 


The  scheme  of  the  work  is  as  follows 


0 
200 
200 
200 
400 


60 
860 

60 
920 

60 

980 

_60 

1040 


3 

1043 

3 

1046 

3 

1049 

3 

1052 


0.8 


1052.8 

0.8 

1053.6 

0^ 

1054.4 

0^ 

1055.2 


-3 

40000 

39997 

80000 

119997 


55200 

346797 

58800 

405597 


3129 
408726 

3138 
411864 

3147 
415011 


842.2 


415853.2 
842.9 

416696.1 
843.5 

417539.6 


+  11 

7999400 

7999411 

23999400 

31998811 


-  4842624131 
1599882200 

-  3242741931 


200 
600 

120000 
239997 

17495820 
49494631 

200 

800 

51600 
291597 

20807820 
70302451 

2969677860 
-  273064071 


214585887 
-58478184 


1226178 
71528629 

1235592 
72764221 


58477522.9 
-661.1 


332682.6 
73096903.6 

333356.9 
73430260.5 


661.1 


=  0.000009 +. 


73430260.5 

a;  =  263. 800009 


618.    Any  root  of  a  number  can  be  extracted  by  Horner's 
Method.     Find  the  fourth  root  of  473. 

Here  aj*  =  473, 

or  x4  + 0x3 +  0x2  + Ox -473  =  0, 

and  x  =  4.66353+. 

If  the   number  be  a  perfect  power,  the  root  will  be 
obtained  exactly. 


HIGHER   NUMERICAL   EQUATIONS.  499 

619.  From  the  preceding  sections  are  derived  the  fol- 
lowing general  directions  for  solving  a  higher  numerical 
equation  whose  roots  are  real  : 

I.    Remove  commensurable  roots  by  §  603. 
II.    By  inspection  and  Descartes'  Rule  find  roughly  the 
situations  of  the  incommensurable  roots. 

III.  Find  one  or  more  figures  of  each  incommensurable 
root  by  §§  607-609. 

IV.  Apply  Horner's  Method  as  set  forth  in  §§  612-618. 

620.  Compute  one  root  of  each  of  the  following  equa- 
tions, carrying  each  result  to  six  decimal  places.  The  last 
two  or  three  places  may  be  found  as  in  the  last  part  of 
§  613.  The  root  to  be  sought  lies  between  the  two  numbers 
placed  in  the  parenthesis  following  each  equation,  except 
in  examples  8-12. 

Exercise  147. 

1.  a;3-f  lOx^-f  6ic-120  =  0.     (2,3.) 

2.  x^-\-x^-\-x  — 100  =  0.     (4,  5.) 

3.  a;*-2x3  +  21a:-23  =  a;.     (1,2.) 

4.  x*-5a:^  +  3a;2  +  35a;-70==0.     (2,3.) 

5.  x^-12^2_|_i2a:-3  =  0.     (-3,-4.) 

6.  x'-\-2x'-\-Sx'-\-4:X^-\-5x-5^S21  =  0.     (8,9.) 

7.  x'  —  50x'-\-S4:0  =  0.     (4,  5.) 

8.  a;3  — 35499  =  0.  11.  cc*  — 147008443  =  0. 

9.  ic^  — 242970624  =  0.     12.  a;^- 551791  =  0. 

10.  ic^  — 707281=0.  13.  ir2_  17^4- 70.3  =:zO.     (7,8.) 

14.  x^-\-9x^-{-2^x-j-17  =  0.     (-4,-5.) 

15.  x*  —  Sx^'}-Ux'-{-4:X-S  =  0.     (0,-1.) 


500  ALGEBRA. 

621.  Many  methods  have  been  found  of  abbreviating 
the  computation,  when  it  is  desired  to  get  the  root  to  a 
large  number  of  decimal  places.  We  give  one  simple  way 
which  can  sometimes  be  profitably  used  after  three  or  four 
decimal  places  have  been  computed. 

Consider  the  example  in  §  613.  If  that  computation  had 
not  been  abbreviated,  the  last  transformed  equation  would 
have  been 

w^  — 1.731  w^  —  8.001213?^  +  0.000899967  =  0.  (1) 

We  have  found  that  w  is  between  0.00011  and  0.00012. 
Computing  the  value  of  w'^  — 1.731 1/;^  with  these  values  of 
w,  and  substituting  the  results  in  (1),  we  have 

8.001213  w  =  0.000899946  ^,  (2) 

and  8.001213  w  =  0.000899942  ^.  (3) 

From  (2),  7^;  =  0.0001124762=^.  (4) 

From  (3),  ?^;=:  0.0001124757=^.  (5) 

Since  w  is  about  one-fourth  the  way  from  0.00011  to 
0.00012,  take  a  value  one-fourth  the  way  from  the  one  in 
(4)  to  that  in  (5),  and 

^  =  0.0001124761^. 

If  we  wish  it  more  accurately  still,  w^  — 1.731?^^  can  be 
recomputed  with  the  value  of  w  just  found ;  and  substitu- 
tion in  (1)  will  give  a  value  of  w  extending  a  few  decimal 
places  further  ;  and  so  on. 

622.  When  all  but  two  roots  of  any  equation  have  been 
determined,  §  576  enables  us  to  get  both  these  roots  at 
once. 

Take  the  equation 

x^-6x^-\-3x-\-5  =  0, 

one  root  of  which  is  1.423112.    Let  this  root  be  represented 
by  a,  and  the  other  two  roots  by  b  and  c. 


HIGHER    NUMERICAL    EQUATIONS.  501 

By  §  576,  a+b  +  c  =  6,  and  abc=  —  5. 

.:  6+ c  =  4.576888, 
bc=  -  3.513427, 
62  4- 2  6c +  c2  =  20.94790, 

4  6c=  -14.05371, 
62  -  2  6c +  c2  =  35.00161, 
6- c=  5.916216, 
26=  10.493104, 
2c  =  -  1.339328, 
6=5.246552, 
and  c  =  -  0.669664. 

The  above  solution  involves  little  labor  when  the  com- 
putation is  made  with  the  aid  of  logarithms. 
Two  roots  of 

x^  —  Ux^-^79x^  —  U0x-{-58  =  0 

are  0.58579  and  8.64575  ;  find  the  others. 


Solutions  of  Special  Forms, 
recurring  equations. 

623.   A  Recurring  Equation,  or  Reciprocal  Equation,  is 

one  in  which  the  coefficients  of  terms  equally  distant  from 
the  extremes  are  numerically  equal ;  the  signs  of  the  cor- 
responding terms  are  either  all  alike  or  all  unlike. 
Below  are  examples  of  recurring  equations  : 

x'-Sx*-\-5x^  —  5x^-\-3x-l=0, 
a*  — 7a:3+ 15a:2  -  7.r  + 1  =0, 
a:«-3a;^-f  5^^-5x^  +  3^-1  =  0. 

The  last  equation  shows  that  when  the  degree  of  the 
equation  is  even,  and  the  signs  of  the  corresponding  terms 
unlike,  the  middle  term  is  wanting.  By  definition,  the 
middle  term  of  such  an  equation  is  both  positive  and  nega- 
tive, and  therefore  must  be  zero. 


502  ALGEBKA. 

624.  If  ?i  he  a  root  of  a  recurring  equation,  -  is  also  a 

a 

root.     For,  in  order  to  obtain  an  equation  whose  roots  are 

reciprocals  of  those  of  the  given  equation,  we  simply  write 
the  coefficients  in  reverse  order  (§  598) ;  this,  however,  does 
not  alter  a  recurring  equation  (except  to  change  all  its 

signs  when  in  one  form),  and  so,  if  a  is  a  root  of  it,  -  is 
also  a  root. 

Recurring  equations  are  called  Eeciprocal  Equations, 
because  of  this  property. 

625.  A  recurring  equation  of  an  odd  degree  has  for  one 
root  —  1  or  -\- 1,  according  as  the  signs  of  the  corresponding 
terms  are  alike  or  unlike. 

x^  +  Ax""-^  +  ^a;«-2  + -izBx'^Ax^l^^^  (1) 

is  a  typical  recurring  equation,  and  may  be  written  in  the 
form 

(ir"±l)  +  ^(x"-id=£c)  +  ^(a;«-2±£c2)+ =  0.        (2) 

When  the  signs  of  the  corresponding  terms  are  alike,  we 
consider  the  upper  signs  in  (2),  since  the  coefficient  a?"  is 
positive.  As  n  is  an  odd  number,  the  substitution  of  — 1 
for  X  will  satisfy  the  equation,  because  the  expression  in 
each  parenthesis  becomes  zero.  Likewise,  when  the  signs 
of  the  corresponding  terms  are  unlike,  we  must  consider 
the  lower  signs  in  (2).  The  value  +1  substituted  for  x 
will  then  satisfy  the  equation. 

626.  A  recurring  equation  of  an  even  degree  has  + 1  Cind 
—  1  as  roots,  when  the  corresponding  terms  have  unlike  signs. 

Equation  (2)  of  §  625  may  be  written,  when  the  corre- 
sponding terms  have  unlike  signs,  as  follows  : 

(x^  —  l)^Ax  (^"-2  —  1)  +  Bx^  (x^-^  —  1)  + =  0. 


HIGHER   NUMERICAL    EQUATIONS.  503 

In  this  case  n  is  an  even  number,  and  every  exponent  of 
X  is  even.  Each  parenthesis  is  therefore  divisible  by  x-—l, 
and  the  whole  equation  is  divisible  by  (x -\-l)  (x—1); 
and  consequently  + 1  and  —  1  are  roots. 

627.  A  recurring  equation  of  an  even  degree  is  reducible 
to  an  equation  of  half  its  degree,  when  the  corresponding 
terms  have  like  signs. 

The  first  and  last  terms  are  positive,  but  the  interme- 
diate terms  need  not  be.  For  convenience,  we  shall  repre- 
sent them  all  as  positive,  and  write  the  equation 

a;2«  +  J^2,.-i_|_j^^2«-2_^ J^Bx^+Ax-\-l=0.  (1) 

Dividing  (1)  by  cc",  and  rearranging  it,  we  have 

(-"+i.)+^(-"-'+,-^.)  +  «(-"-^+i^.)+ =0.(2) 

The  middle  term  of  (1),  since  it  is  of  the  form  Ex"",  will 
appear  as  a  known  number  {E)  in  (2). 

^^^  ('^  +  ^)  =  ^^- 

Then  (xa -f- ^^  =  (x  4- -^-2  =  y2_2  ; 

(xi  +  —^  =  (x^  +  ^\  —2  =  i/*-iy''+2;     and  so  on. 


In  general, 


(^"  +  ^)=2/"-^y"-' 


Hence,  (2)  may  be  expressed  in  terms  of  y,  involving  no 
power  higher  than  the  nth,  and  being,  therefore,  of  half  the 
degree  of  (1). 


504  ALGEBRA. 

628.  Any  recurring  equation  is  reducible  to  one  of  half 
its  degree. 

For  the  equations  of  §  &25,  being  divided  respectively 
by  ic  + 1  and  x  —  1,  are  reduced  to  the  form  of  §  627  ;  the 
equation  of  §  626,  being  divided  by  x^  —  1,  is  reduced  to 
the  form  of  §  627.  These  reduced  equations  can  now  be 
treated  by  the  method  of  §  627. 

Solve     £c^  —  llcc^  +  17a;^  +  17a;2  — 110^4-1  =  0. 

One  root  is  —  1,  and  dividing  the  equation  by  x  +  1,  we  have 

x*  -  12  ic3  +  29x2  —  12  X  +  1  =  0. 
Divide  by  x^,  rearrange  the  terms,  and  add  2  to  both  sides. 

Then  (x  + -V- 12  ^x  + -)  +  29  =  2; 

whence  (x+  -)  =  9or3, 

and  X  =  i  (9  ±  V77)  or  i  (3  ±  Vs). 

i  (9  +  V77)  is  the  reciprocal  of  i  (9  -  V77),  and  i  (3  +  VH)  is  the 
reciprocal  of  ^  (3  —  V5),  the  product  in  each  case  being  unity. 

Exercise  148. 
Solve  : 

1.  x^-\-lx^  —  lx  —  l  =  0. 

2.  x^-{-x^-\-x^-\-x-\-l={). 

3.  x^  —  ^x'  +  ^x^  —  ^x^^^x  —  l  =  Q^, 

4.  a;^  — 5x3  +  6rz;^  — 5cc  +  l  =  0. 

5.  2ic^-5a;«  +  6^2_5^_|_2=:0. 

6.  x'  —  4.x^-\-x^-\-x^  —  i:X-\-l=0. 

7.  x*-10x3  +  26x2-10x'  +  l  =  0. 

8.  x^  -\-  mx^  +  mx  -|- 1  =  0. 

9.  x'^\.  =  ^. 

10.    3cc^-2x*  +  5x3-5a^2^2a;-3  =  0. 


higher  numerical  equations.  605 

Exponential  Equations. 

629.  An  Exponential  Equation  is  one  in  which  an  un- 
known number  appears  as  an  exponent. 

I.  Solve  a^z=m.  Taking  the  logarithms  of  both  mem- 
bers, we  have  x  log  a^log  m:  whence  x^^ — 

log  a 

II.  Solve  £c^  =  m.  As  before,  x  log  x  =  log  m.  Find  by- 
trial,  from  a  logarithm  table,  a  value  of  x  which  satisfies 
the  equation.     Thus,  in  the  equation  x^=-l : 

By  trial,  2.3  log  2.3     =  0.83197  ; 

2.32  log  2.. 32    =0.84793; 
2.316  log  2.31G  =  0.84474.     Hence,  x  =  2.316. 

Exercise  149. 
Solve : 

1.  11^  =  346.  5.  4^  =  3.74.  8.  14.74^  =  8.64. 

2.  3^  =  10.  6.  146-^  =  12984.  9.  a:^  =  2.767. 

3.  10^  =  745.  7.  0.2-^  =  0.4.  10.  a^^  =  23.10. 

4.  7^  =  324. 

In  Compound  Interest,  if  P  denotes  the  principal,  A  the 
amount,  r  the  rate,  t  the  integral  number  of  years  (or  other 
periods  of  time),  we  have  the  formula, 


A  =  P(1 

^ry, 

whence 

log^ 
log( 

A 
P' 

-logP 

dllU. 

:i+^) 

Find  t  in 

the 

following  cases : 

11. 

P-- 

=  750,          A  = 

=  1797.42, 

r 

=  6%. 

12. 

P-- 

=  780,          A  = 

=  1559.22, 

r 

=  8%. 

13. 

P-- 

=  5630.75,  ^  = 

=  21789.22, 

r 

=  7%- 

14. 

P-- 

=  300,         A  = 

=  515.46, 

r 

=  7%. 

15. 

P-- 

=  84.65,       A  = 

=  289.47, 

r 

=n% 

606  ALGEBRA. 

Cardan's  Method  for  Cubic  Equations. 
630.   The  general  form  of  a  cubic  equation  is 

x^  -\-  mx^  -\-7ix-\-c^0.  (1) 

Let  x  =  y  —  --  (2) 

By  §  GOl,  equation  (1)  reduces  to  the  form 

2/3  +  p?/  +  g  =  0.  (3) 

Let  y=^-A.  (4) 

Substitute  this  value  of  y  in  (3)  and  reduce : 

Vr)3         «2 
27  "^  4  ' 


and  ^=\-^'?±Vi+4- 


(5) 


From  (2)  and  (4),    ^J  =  ^  "  ^  "  3  *  («) 

When  ^  has  been  determined  from  (5),  x  can  be  found 
from  (6).  In  solving  these  equations,  it  is  best  to  get 
from  (5)  the  value  of  z  most  easily  obtained.  Then  find 
the  value  of  x  from  (6),  and  designate  this  value  by  r ; 
then  divide  (1)  by  x  —  r,  and  solve  the  resulting  quadratic. 
Or  the  method  of  §  622  may  be  employed.  If  the  original 
equation  does  not  contain  the  square  of  the  unknown  num- 
ber, it  will  be  in  the  form  of  (3),  and  the  value  of  y  can 
be  found  from  (5)  and  (4). 

Solve  x^  —  ^x''  —  12x-\-112  =  Q, 

By(2),  x  =  y-^=y  +  2. 

Substitution  gives  y^  —  24  y  +  72  =  0. 
Here  p  =  —  24,     and     q  =  72. 


HIGHER    NUMERICAL    EQUATIONS.  507 


= 

<l- 

-  36  ±  ^- 

13824 

27 

+ 

5184 
4 

rz 

r- 

T  36  ±  28  = 

-2. 

•.  from  (5), 


From  (6),  x=-^.')  ^  t^ 

Divide  the  original  equation  by  x  +  4  ;  the  resulting  equation  is 
x2  -  10  x  +  28  =  0 ; 
whence  x  =  5  ±  V—  3. 

631.   If  j9  is  negative,  and  p  and  q  have  such  values 

that  £^  + 1;  is  negative,  equation  (5)  shows  that  Cardan's 

Method  is  unsatisfactory,  since  z  is  the  cube  root  of  an 
expression  which  is  partly  imaginary,  and  we  have  no 
general  rule  for  extracting  such  a  root.  This  case,  which 
is  called  the  irreducible  case,  can  be  solved  by  Trigonometry, 
and  may  be  illustrated  by  an  example. 

Solve  'if  —  15u  —  4.  =  0. 

Substitution  in  (5)  gives 


=^/-^/^ 


+  4=  V2±iiv=n. 

It  is  possible  to  determine,  by  trial,  that 

(2  ±  V^)3  =  2  ±  11 V^  ; 
.-.  z  =  2  ±  V^, 
and,  from  (4),  y  =  4. 

Divide  the  given  equation  by  y  —  4,  and  solve  the  resulting  quad- 
ratic ;  then 

y  =  -  2  ±  V3. 

golyg  .  Exercise  150. 

1.  a;''+12a;'^  +  45x  +  50  =  a.  5.  2/«  +  48 ?/ +  504  =  0. 

2.  x3  —  21a;2+159ic  — 490  =  0.  6.  if  —  21y  —  SU  =  0. 

3.  x^  —  Q>x''-^13x-10  =  0.  7.  ?/3  — 37/  +  2==0. 

4.  a:»  +  3x-  +  9a^  — 13  =  0.  8.  f  —  QOy-{-Q>ll  =  0. 


608  ALGEBRA. 

Trigonometric  Solution  of  Cubic  Equations. 

632.  Trigonometric  formulas  are  used  in  solving  quad- 
ratic and  cubic  equations  ;  and  we  subjoin  those  for  cubics, 
so  that  the  student  may  use  them  when  he  has  acquired  a 
knowledge  of  Trigonometry. 

In  the  formulas,  p  is  essentially  positive,  but  q  may  be 
either  positive  or  negative.     We  distinguish  three  cases  : 

I.    oc?  -\-px  -}-  2'  =  0  ; 
11.    x^—x,x^q  =  0,  and  |y< j; 

III.   x^-px-^q  =  0,  and  ^>|'-    • 

The  formulas  for  Case  I.  are  as  follows : 

tan^=^-i/^-  tan  i  0  =  Vtan  i  ^. 

Xx=  —2  -J|  cot  0. 

^2  =  -v/'l  cot  0  +  CSC  0  Vp  V^^. 
Xs  =  a/^  cot  (f>  —  CSC(l)  Vp  V—  1. 

The  formulas  for  Case  II.  are  as  follows : 

sme  =  -^-J^'  tan  i  0  =  Vtan  i  ^. 

xi  =  -  2  -i/|  CSC  0. 
X2  =  -v/^  CSC  0  +  cot  0  Vp  V—  1. 
Xs  =  -1/ ^  CSC  0  —  cot  0  Vp  V^. 


HIGHER    NUMERICAL    EQUATIONS. 

The  formulas  for  Case  III.  are  as  follows  : 


509 


\p  \p 

sin  ^  6. 


X2=2^|sin(G0°-i5). 
JC8=-2^|sin(60°  +  i^), 


If  sin  ^  or  tan  B  is  positive,  0  is  in  the  first  quadrant ;  if 
it  is  negative,  Q  is  in  the  fourth  quadrant. 

In  the  preceding  formulas,  x^,  Xg,  x^  denote  the  three 
roots  in  each  case. 

633.    Take  the  difficult  equation 
which  comes  under  Case  III. 


log  403  =  2.60531 

e  =  68°  32'  0" 

log  441  =  2.64444 

ie=22°50'40'' 

log    46  =  1.66276 

logsini^=  9.58909 -10 

log  147  =  2.16732 
log     2  =  0.30103 

log2-J|  =  0.04291 

log     p  =  0.96087 - 

-10 

log  sin 

i(60°-i^)  =  9.78102-10 

log     3  =  0.47712 

log  sin 

L  (60°  -\-ie)  =  9. 99660  —  10 

log     q  =  9.49544  - 

-10 

logXi=  9.63200 -10 

log  3^  =  9.97256- 

-10 

logx2=  9.82393 -10 

colog  2i)=  9.73810- 

,         /3 

-10 

log  (-X3)  =  0.03951 
Xi  =  0.42855 

log^-=  0.25812 

X2  =  0.66670 

logsin^=  9.96878 - 

-10 

Xs=-  1.09525 

510  ALGEBRA. 

This  is  an  unfavorable  example,  because  sin  6  is  large, 
and  9  is  therefore  inaccurately  determined ;  yet  the  values 
of  the  roots  are  near  the  true  values,  which  are  0.42857, 

0.66667,  and  —1.09524.     If  log  J5,  which  was  0.258125, 

had  been  assumed  to  be  0.25813,  the  answers  would  have 
been  0.42857,  0.66668,  and  —1.09525. 

Exercise  151. 
Solve  : 

1.  aj-'^  +  Sx  — 5  =  0.  3.  x^  —  lx-{-ll  =  0. 

2.  x^+7cc  +  3  =  0.  4.  x^  —  ^x  —  6  =  0. 

5.  X'"  — 5a;  +  4  =  0. 


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